MCQ

MCQ 11 Mark

If the equation $\frac{\lambda(\text{x}+1)^2}{3}+\frac{(\text{y}+2)^2}{4}=1$ represents a circle then $\lambda$:

A
$1$
✓
$\frac{3}{4}$
C
$0$
D
$-\frac{3}{4}$

Answer

Correct option: B.

$\frac{3}{4}$

$\frac{3}{4}$

Solution:
$\frac{\lambda(\text{x}+1)^2}{3}+\frac{(\text{y}+2)^2}{4}=1$
for a circle of a $(\text{x}-\alpha)^2+6 (\text{y}-\beta)^2=1$ then (a = 6)
$\frac{\lambda}{3}=\frac{1}{4}$
$\lambda=\frac{3}{4}$

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MCQ 21 Mark

The eccentricity of the ellipse, if the distance between the foci is equal to the length of the latus-rectum, is:

✓
$\frac{\sqrt{5}-1}{2}$
B
$\frac{\sqrt{5}+1}{2}$
C
$\frac{\sqrt{5}-1}{4}$
D
$\text{none of these}$

Answer

Correct option: A.

$\frac{\sqrt{5}-1}{2}$

According to the question, the distance between the foci is equal to the length of the latus rectum.
$\frac{2\text{b}^2}{\text{a}}=2\text{ae}$
$\Rightarrow\text{b}^2=\text{a}^2\text{e}$
Now, $\text{e}=\sqrt{1-\frac{\text{b}^2}{\text{a}^2}}$
$\Rightarrow\text{e}=\sqrt{1-\frac{\text{a}^2\text{e}}{\text{a}^2}}$
$\Rightarrow\text{e}=\sqrt{1-\text{e}}$
On squaring both sides, we get:
$\text{e}^2\text{e}-1=0$
$\Rightarrow\text{e}=\frac{-1\pm\sqrt{1+4}}{2}$
$\Rightarrow\text{e}=\frac{\sqrt{5}-1}{2}$ $(\because\text{e}\text{ cannot be negative}\big)$

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MCQ 31 Mark

Find the equation of the circle. Centered at (3, -2) with radius 4:

A
$x^2+y^2+6 x-4 y=3$
✓
$x^2+y^2-6 x+4 y=3$
C
$x^2+y^2-3 x+2 y=-3$
D
$x^2+y^2+3 x-2 y=-3$

Answer

Correct option: B.

$x^2+y^2-6 x+4 y=3$

$x^2+y^2-6 x+4 y=3$

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MCQ 41 Mark

The equation of ellipse whose one focus is at (4, 0) and whose eccentricity is $\frac{4}{5}$ is:

A
$\frac{\text{x}^2}{5}+\frac{\text{y}^2}{9}=1$
✓
$\frac{\text{x}^2}{25}+\frac{\text{y}^2}{9}=1$
C
$\frac{\text{x}^2}{9}+\frac{\text{y}^2}{5}=1$
D
$\frac{\text{x}^2}{9}+\frac{\text{y}^2}{25}=1$

Answer

Correct option: B.

$\frac{\text{x}^2}{25}+\frac{\text{y}^2}{9}=1$

$\frac{\text{x}^2}{25}+\frac{\text{y}^2}{9}=1$

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MCQ 51 Mark

The equation of the incircle formed by the coordinate axes and the line 4x + 3y = 6 is:

A
$x^2+y^2-6 x-6 y+9=0$
✓
$4\left(x^2+y^2-x-y\right)+1=0$
C
$4\left(x^2+y^2+x+y\right)+1=0$
D
None of these

Answer

Correct option: B.

$4\left(x^2+y^2-x-y\right)+1=0$

$4\left(x^2+y^2-x-y\right)+1=0$

Solution:
The line 4x + 3y = 6 cuts the coordinate axes at $\Big(\frac{3}{2},\ 0\Big)$ and (0, 2)
The coordinates of the incentre is $\Big(\frac{\text{ax}_1+\text{bx}_2+\text{cx}_3}{\text{a+b+c}},\ \frac{\text{ay}_1+\text{by}_2+\text{cy}}{\text{a+b+c}}\Big)$
Here, $\text{a}=\frac{5}{2},\ \text{b}=\frac{3}{2},\ \text{c}=2,\ \text{x}_1=0,\ \text{y}_1\\=0,\ \text{x}_2=0,\ \text{y}_2=2,\ \text{x}_3=\frac{3}{2},\ \text{y}_3=0$
Thus, the coordinates of the incentre:
$\Big(\frac{0+0+3}{6},\ \frac{0+3+0}{6}\Big)$
$=\big(\frac{1}{2},\ \frac{1}{2}\Big)$
The equation of the incircle:
$\Big(\text{x}-\frac{1}{2}\Big)^2+\Big(\text{y}-\frac{1}{2}\Big)^2=\text{a}^2$
Also, radius of the incircle $=\frac{\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}}{\text{s}}$
Here, $\text{s}=\frac{\text{a+b+c}}{2}=\frac{\frac{5}{2}+\frac{3}{2}+2}{2}=3$
$\therefore$ Radius of the incircle $=\sqrt{\frac{3(3-\text{a})(3-\text{b}(3-\text{c}))}{3}}$
$=\frac{\sqrt{3\Big(3-\frac{5}{2}\Big)\Big(3-\frac{3}{2}\Big)(3-\text{c})}}{3}$
$=\frac{\sqrt{3\Big(3-\frac{1}{2}\Big)\Big(\frac{3}{2}\Big)}}{3}$
$=\frac{1}{2}$
The equation of circle:
$\Big(\text{x}-\frac{1}{2}\Big)^2+\Big(\text{y}-\frac{1}{2}\Big)^2=\frac{1}{4}$
$\Rightarrow4(\text{x}^2+\text{y}^2)-\text{x}-\text{y}+1=0$

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MCQ 61 Mark

If the centroid of an equilateral triangle is (1, 1) and its one vertex is (-1, 2), then the equation of its circumcircle is:

✓
$x^2+y^2-2 x-2 y-3=0$
B
$x^2+y^2+2 x-2 y-3=0$
C
$x^2+y^2+2 x+2 y-3=0$
D
None of these

Answer

Correct option: A.

$x^2+y^2-2 x-2 y-3=0$

$x^2+y^2-2 x-2 y-3=0$

Solution:

The centre of the circumcircle is (1, 1).
Radius of the circumcircle
$\therefore$ Equation of the circle: $=\sqrt{(1+1)^2+(1-2)^2}=\sqrt{5}$
$(x-1)^2+(y-1)^2=5$
$\Rightarrow x^2+y 2-2 x-2 y-3=0$

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MCQ 71 Mark

The line $2x - y + 4 = 0$ cuts the parabola $y^2 = 8x$ in $P$ and $Q$. The mid-point of $PQ$ is

A
(1, 2)
B
(1, -2)
✓
(-1, 2)
D
(-1, -2)

Answer

Correct option: C.

(-1, 2)

(-1, 2)

Solution:
Let the coordinates of P and Q be $\left(a \mathrm{t}_1{ }^2, 2 \mathrm{a} \mathrm{t}_1\right)$ and $\left(\mathrm{at}_2{ }^2, 2 \mathrm{a} \mathrm{t}_2\right)$, respectively.
Slope of $\mathrm{PQ}=\frac{2 \mathrm{at}_2-2 \mathrm{at}}{\mathrm{at}_2^2-\mathrm{at}_1^2} \ldots$(1)
But, the slope of $P Q$ is equal to the slope of $2 x-y+4=0$.
$\therefore$ Slope of $\mathrm{PQ}=\frac{-2}{-1}=2$
From (1),
$\frac{2 \mathrm{at}_2-2 \mathrm{a} t_1}{\mathrm{at}_2^2-\mathrm{at}_1^2}=2 \ldots$
Putting $4 a=8$,
$a=2$
$\therefore$ Focus of the given parabola $=(a, 0)=(2,0)$
Using equation (2):
$\frac{4\left(t_2-t_1\right)}{2\left(t_2^2-t_1^2\right)}=2$
$\frac{\left(t_2-t_2\right)}{\left(t_2^2-t_1^2\right)}=1$
$\Rightarrow t_1+t_2=1$
As, points $P$ and $Q$ lie on $2 x-y+4=0$
$\Rightarrow P\left(a t_1{ }^2, 2 a t_1\right) \text { or } P\left(2 t_1{ }^2, 4 t_1\right) \text { lie on line } 2 x-y+4=0$
$\Rightarrow 2\left(2 t_1{ }^2\right)-\left(4 t_1\right)+4=0$
$\Rightarrow t_1{ }^2-t_1+1=0 \ldots(3)$
Also, $Q\left(a t_2{ }^2, 2 a t_2\right)$ or $P\left(2 t_2{ }^2, 4 t_2\right)$ lie on line $2 x-y+4=0$
$\Rightarrow 2\left(2 t_2^2\right)-\left(4 t_2\right)+4=0$
$\Rightarrow t_2^2-t_2+1=0 \ldots(4)$
Adding (3) and (4), we get,
$\Rightarrow t_1^2-t_1+1+t_2{ }^2-t_2+1=0$
$\Rightarrow\left(t_1^2+t_2^2\right)-\left(t_1+t_2\right)+2=0$
$\Rightarrow\left(t_1^2+t_2^2\right)-1+2=0\left[t_1+t_2=1, \text { proved above }\right]$
$\Rightarrow\left(t_1^2+t_2{ }^2\right)=-1$
Let $\left(x_1, y_1\right)$ be the mid-point of PQ.
Then, we have:
$\mathrm{y}_1=\frac{2 \mathrm{a} t_2+2 \mathrm{a} t_1}{2}=2\left(\mathrm{t}_1+\mathrm{t}_2\right)=2$
$\text { And, } \mathrm{x}_1=\frac{a \mathrm{t}_1^2+a \mathrm{t}_2^2}{2}=\mathrm{t}_1^2+\mathrm{t}_2^2=-1$
$\Rightarrow\left(\mathrm{x}_1, \mathrm{y}_1\right)=(-1,2)$

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MCQ 81 Mark

The circle $x^2+y^2-3 x-4 y+2=0$ cuts x-axis:

A
(2, 0), (-3, 0)
B
(3, 0), (4, 0)
C
(1, 0), (-1, 0)
✓
(1, 0), (2, 0)

Answer

Correct option: D.

(1, 0), (2, 0)

(1, 0), (2, 0)

Solution:
$x^2+y^2-3 x-4 y+2=0$
$x \text {-axis will be cut when } y=0$
$\text { put } y=0$
$x^2-3 x+2=0$
$(x-2)(x-1)=0$
$x=1,2$
$\text { points }(1,0),(2,0)$

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MCQ 91 Mark

The equation of the circle passing through (3, 6) and whose centre is (2, -1) is:

✓
$x^2+y^2-4 x+2 y=45$
B
$x^2+y^2-4 x-2 y+45=0$
C
$x^2+y^2+4 x-2 y=45$
D
$x^2+y^2-4 x+2 y+45=0$

Answer

Correct option: A.

$x^2+y^2-4 x+2 y=45$

$x^2+y^2-4 x+2 y=45$

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MCQ 101 Mark

Which of the following equations of a circle has center at $(1,-3)$ and radius of $5$ :

A
$x^2+y^2=25$
✓
$(x-1)^2+(y+3)^2=25$
C
$(x-1)^2+(y-3)^2=25$
D
$(x+1)^2+(y-3)^2=25$

Answer

Correct option: B.

$(x-1)^2+(y+3)^2=25$

$(x-1)^2+(y+3)^2=25$

Solution:
The general equation of a circle with center at $(a, b)$ and radius $r$ is $(x-a)^2+(y-b)^2=r^2$
So substituting the values we get the circle equation as $(x-1)^2+(y+3)^2=25$

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MCQ 111 Mark

The eccentricity of the conic $9\text{x}^2+25\text{y}^2=225$ is:

A
$\frac{2}{5}$
✓
$\frac{4}{5}$
C
$\frac{1}{3}$
D
$\frac{1}{5}$

Answer

Correct option: B.

$\frac{4}{5}$

$9\text{x}^2+25\text{y}^2=225$
$\Rightarrow\frac{\text{x}^2}{25}+\frac{\text{y}^2}{9}=1$
Comparing it with $\Rightarrow\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1,$ we get:
$\text{a}=5$ and $\text{b}=3$
Here, a > b, so the major and the minor axes of the ellipse are along the x−axis and y−axis, respectively.
Now, $\text{e}=\sqrt{1-\frac{\text{b}^2}{\text{a}^2}}$
$\Rightarrow\text{e}=\sqrt{1-\frac{9}{25}}$
$\Rightarrow\text{e}=\sqrt{\frac{16}{25}}$
$\Rightarrow\text{e}=\frac{4}{5}$

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MCQ 121 Mark

The equation of the parabola whose vertex is $(a, 0)$ and the directrix has the equation $x+y=3 a$, is

A
$x^2+y^2+2 x y+6 a x+10 a y+7 a^2=0$
✓
$x^2-2 x y+y^2+6 a x+10 a y-7 a^2=0$
C
$x^2-2 x y+y^2-6 a x+10 a y-7 a^2=0$
D
None of these

Answer

Correct option: B.

$x^2-2 x y+y^2+6 a x+10 a y-7 a^2=0$

$x^2-2 x y+y^2+6 a x+10 a y-7 a^2=0$

Solution:
Given:
The vertex is at (a, 0) and the directrix is the line x + y = 3a.
The slope of the line perpendicular to x + y = 3a is 1.
The axis of the parabola is perpendicular to the directrix and passes through the vertex.
$\therefore$ Equation of the axis of the parabola = y − 0 = 1(x - a) ...(1)
Intersection point of the directrix and the axis is the intersection point of (1) and x + y = 3a.
Let the intersection point be K.
Therefore, the coordinates of K are (2a, a)
The vertex is the mid-point of the segment joining K and the focus (h, k).
$\therefore\ \text{a}=\frac{2\text{a+h}}{2},\ 0=\frac{\text{a+k}}{2}$
h = 0, k = -a
Let P (x, y) be any point on the parabola whose focus is S (h, k) and the directrix is x + y= 3a.

Draw PM perpendicular to x + y = 3a.
Then, we have:
$SP = PM$
$\Rightarrow SP^2 = PM^2$
$\Rightarrow\ (\text{x}-0)^2+(\text{y+a})^2=\Big(\frac{\text{x+y}-3\text{a}}{\sqrt2}\Big)^2$
$\Rightarrow\ \text{x}^2+(\text{y+a})^2=\Big(\frac{\text{x+y}-3\text{a}}{\sqrt2}\Big)^2$
$\Rightarrow\ 2\text{x}^2+2\text{y}^2+2\text{a}^2+4\text{ay}=\text{x}^2+\text{y}^2+9\text{a}^2+2\text{xy}-6\text{ax}-6\text{ay}$
$\Rightarrow\ \text{x}^2+\text{y}^2-7\text{a}^2+10\text{ay}+6\text{ax}=0$

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MCQ 131 Mark

The equation $\sqrt{(\text{x}-2)^{2}+\text{y}^{2}}+\sqrt{(\text{x}+2)^{2}+\text{y}^{2}}=5$ represents:

A
A circle
✓
Ellipse
C
Line segment
D
An empty set

Answer

Correct option: B.

Ellipse

let A (2, 0), B (-2, 0) and P (x, y) be three points AB = 4
Given: that, $\sqrt{(\text{x}-2)^{2}+\text{y}^{2}}+\sqrt{(\text{x}+2)^{2}+\text{y}^{2}}=5>\text{AB}$
⇒ PA + PB = constant > AB
$\therefore$ locus of P is an ellipse.

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MCQ 141 Mark

The equation of the circle which touches the axes of coordinates and the line $\frac{\text{x}}{3}+\frac{\text{y}}{4}=1$ and whose centres lie in the first quadrant is $x^2 + y^2 − 2cx − 2cy + c^2 = 0$, where c is equal to:

A
4
B
2
C
3
✓
6

Answer

Correct option: D.

Solution:
The equation of the circle that touches the axes of coordinates is $x^2 + y^2 - 2cx − 2cy + c^2 = 0$.
Also, $x^2 + y^2 − 2cx − 2cy + c^2 = 0$ touches the line $\frac{\text{x}}{3}+\frac{\text{y}}{4}=1$ or 4x +3y -12 = 0.
Since the circle lies in the first quadrant, it centre is is (c, c).

From the figure, we have:
$\Bigg|\frac{4\text{c}+3\text{c}-12}{\sqrt{4^2+3^3}}\Bigg|=\text{c}$
$\Rightarrow\frac{7\text{c}-12}{5}=\text{c}$
$\Rightarrow\text{c}=6$

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MCQ 151 Mark

The intercept on the line $y=x$ by the circle $x^2+y^2-2 x=0$ is $A B$. Equation of the circle with $A B$ as a diameter is:

A
$x^2+y^2+x+y=0$
✓
$x^2+y^2-x-y=0$
C
$x^2+y^2+x-y=0$
D
None of these

Answer

Correct option: B.

$x^2+y^2-x-y=0$

$x^2+y^2-x-y=0$

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MCQ 161 Mark

Choose the correct answer. The eccentricity of the hyperbola whose latus rectum is 8 and conjugate axis is equal to half of the distance between the foci is:

A
$\frac{4}{3}$
B
$\frac{4}{\sqrt{3}}$
✓
$\frac{2}{\sqrt{3}}$
D
none of these.

Answer

Correct option: C.

$\frac{2}{\sqrt{3}}$

Let the equation of the hyperbola be $\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}=1$
Length of latus rectum = 8
$\therefore\ \frac{2\text{b}^2}{2}=8$
$\Rightarrow\text{b}^2=4\text{a}$
Conjugate axis = half of the distance between the foci
$\therefore\ 2\text{b}=\text{ae}$
Now, $\text{b}^2=\text{a}^2(\text{e}^2-1)$
From eqs. (i) and (iii), we get
$\frac{\text{a}^2\text{e}^2}{4}=\text{a}^2(\text{e}^2-1)$
$\Rightarrow\text{e}^2=4\text{e}^2-4$
$\Rightarrow\text{e}^2=\frac{4}{3}$
$\Rightarrow\text{e}=\frac{2}{\sqrt{3}}$

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MCQ 171 Mark

Find the equation of a circle with center (0, 0) and radius 5:

A
$x^2+y^2=5$
B
$x^2-y^2=25$
✓
$x^2+y^2=25$
D
$(x-1)^2+(y+1)^2=25$

Answer

Correct option: C.

$x^2+y^2=25$

$x^2+y^2=25$

Solution:
Compare the equation with the standard form with center at $(\mathrm{h}, \mathrm{k})$ and radius r is.
$(x-h)^2+(y-k)^2=r^2$
Substitute the value of $(h, k)=(0,0)$ and $r=5$.
Then, the equation becomes $x^2+y^2=25$

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MCQ 181 Mark

If the vertex = (2, 0) and the extremities of the latus rectum are (3, 2) and (3, -2), then the equation of the parabola is:

A
$y^2=2 x-4$
B
$x^2=4 y-8$
✓
$y^2=4 x-8$
D
None

Answer

Correct option: C.

$y^2=4 x-8$

$y^2=4 x-8$

Solution:
$y^2=4 a x$
$(y-0)^2=4 a(x-2)$
$y^2=4 a(x-2)$
$y^2=4(x-2)$
$y^2=4 x-8$

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MCQ 191 Mark

The equation of the circle passing through the point $(1,1)$ and having two diameters along the pair of lines $x^2-y^2-$ $2 x+4 y-3=0$, is:

✓
$x^2+y^2-2 x-4 y+4=0$
B
$x^2+y^2+2 x+4 y-4=0$
C
$x^2+y^2-2 x+4 y+4=0$
D
None of these

Answer

Correct option: A.

$x^2+y^2-2 x-4 y+4=0$

$x^2+y^2-2 x-4 y+4=0$

Solution:
Let the required equation of the circle be $(x-h)^2+(y-k)^2=a^2$.
Comparing the given equation $x^2-y^2-2 x+4 y-3=0$ with $a x^2+b y^2+2 h x y+2 g x+2 f y+c=0$, we get:
$a=1, b=-1, h=0, g=-1, f=2, c=-3$
Intersection point $\left(\frac{\mathrm{hf}-\mathrm{bg}}{\mathrm{ab}-\mathrm{h}^2}, \frac{\mathrm{gh}-\mathrm{af}}{\mathrm{ab}-\mathrm{h}^2}\right)=\left(\frac{-1}{-1}, \frac{-2}{-1}\right)=(1,2)$
Thus, the centre of the circle is $(1,2)$
The equation of the required circle is $(x-1)^2+(y-2)^2=a^2$
Since circle passes through ( 1,1 ), we have:
$1=a^2$
$\therefore$ Equation of the required circle:
$(x-1)^2+(y-2)^2=1$
$\Rightarrow x^2+y^2-2 x-4 y+4=0$

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MCQ 201 Mark

If the point $(2, k)$ lies outside the circles $x^2+y^2+x-2 y-14=0$ and $x^2+y^2=13$ then klies in the interval:

A
$(-3,\ -2)\cup(3,\ 4)$
B
$-3,\ 4$
✓
$(-\infty,\ -3)\cup(4,\ \infty)$
D
$(-\infty,\ -2)\cup(3,\ \infty)$

Answer

Correct option: C.

$(-\infty,\ -3)\cup(4,\ \infty)$

$(-\infty,\ -3)\cup(4,\ \infty)$

Solution:
The given equations of the circles are $x^2+y^2+x-2 y-14=0$ and $x^2+y^2=13$.
Since $(2, k)$ lies outside the given circles, we have:
$4+k^2+2-2 k-14>0 \text { and } 4+k^2>13$
$\Rightarrow k^2-2 k-8>0 \text { and } k^2>9$
$\Rightarrow(k-4)(k+2)>0 \text { and } k^2>9$
$\Rightarrow k>4 \text { or } k<-2 \text { and } k>3 \text { or } k<-3$
$\Rightarrow k>4 \text { and } k<-3$
$\Rightarrow k \in(-\infty,-3) \cup(4, \infty)$

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MCQ 211 Mark

The difference between the lengths of the major axis and the latus-rectum of an ellipse is

A
ae
B
2ae
C
$ae^2$
✓
$2 a e^2$

Answer

Correct option: D.

$2 a e^2$

$2 a e^2$

Solution:
Length of the latus rectum $=\frac{2\text{b}^2}{\text{a}}$
and $\text{e}=\sqrt{1-\frac{\text{b}^2}{\text{a}^2}}$
$\text{a}^2\text{e}^2=\text{a}^2-\text{b}^2$
$\Rightarrow\text{b}^2=\text{a}^2-\text{a}^2\text{e}^2$
$\Rightarrow\text{b}^2=\text{a}^2(1-\text{e}^2)$
$\therefore\ $Length of the latus rectum $=\frac{2\text{a}^2(1-\text{e}^2)}{\text{a}}=2\text{a}(1-\text{e}^2)$
Length of the major axis = 2a
Difference between length of latus rectum and length of major axis $=2\text{a}-2\text{a}(1-\text{e}^2)$
$\\=2\text{a}-2\text{a}+2\text{ae}^2\\=2\text{ae}^2$

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MCQ 221 Mark

The circle $x^2+y^2+2 g x+2 f y+c=0$ does not intersect $x$-axis, if:

✓
$\mathrm{g}^2<\mathrm{c}$
B
$g^2>c$
C
$g^2>2 c$
D
None of these

Answer

Correct option: A.

$\mathrm{g}^2<\mathrm{c}$

$\mathrm{g}^2<\mathrm{c}$

Solution:
Given:
$x^2+y^2+2 g x+2 f y+c=0$ ......... (1)
The given circle intersects the $x$-axis.
The equation of circle becomes $x^2+2 \mathrm{gx}+\mathrm{c}=0$ ......... (2)
Solving equation (2):
$\therefore \text { Discriminant, } D=\sqrt{4 g^2-4 c} \geq 0$
$\Rightarrow 4 g^2-4 c \geq 0$
$\Rightarrow g^2-c \geq 0$
$\Rightarrow g^2 \geq c$
Hence, if $\mathrm{g}^2<\mathrm{c}$, then the given circle will not intersect the x -axis.

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MCQ 231 Mark

The vertex of the parabola $(y+a)^2=8 a(x-a)$ is

A
(-a, -a)
✓
(a, -a)
C
(-a, a)
D
None of these

Answer

Correct option: B.

(a, -a)

(a, a)

Solution:
Given:
The equation of the parabola is $(y+a)^2=8 a(x-a)$.
Putting $X=x-a, Y=y+a$
$Y^2=8 a X$
Vertex $=(X=0, Y=0)=(x-a=0, y+a=0)=(x=a, y=-a)$
Hence, the vertex is at $(a, a)$.

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MCQ 241 Mark

The equation $x^2 + y^2 + 2x - 4y + 5 = 0$ represents:

✓
A point.
B
A pair of straight lines.
C
A circle of non-zero radius..
D
None of these.

Answer

Correct option: A.

A point.

A point

Solution:
The radius of the given circle $=\sqrt{1^1+(-2)^2-5}=0$
Hence, the radius of the given circle is zero, which represents a point.

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MCQ 251 Mark

The equation of parabola with vertex at origin and directrix. x - 2 = 0 is:

A
$y^2=-4 x$
B
$y^2=4 x$
✓
$y^2=-8 x$
D
$y^2=8 x$

Answer

Correct option: C.

$y^2=-8 x$

$y^2=-8 x$

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MCQ 261 Mark

What is the approximate radius of the circle whose equation is $(\text{x}-\sqrt{3})^2+(\text{y}+2)^2=11$:

A
1.71
B
2.33
✓
3.32
D
3.85

Answer

Correct option: C.

3.32

The radius of given circle is $\sqrt{11}=3.32$

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MCQ 271 Mark

Find the value of a if $y^2 = 4ax$ pases through (8, 8):

✓
2
B
4
C
8
D
None

Answer

Correct option: A.

Solution:
Given point (8, 8)
Given equation $y^2= 4ax$
$\Rightarrow8^2=4\text{a}(8)$
$64 = 32\text{a}$
$\text{a}=\frac{64}{32}$
$\text{a} = 2$

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MCQ 281 Mark

If $V$ and $S$ are respectively the vertex and focus of the parabola $y^2+6 y+2 x+5=0$, then $S V=$

A
$2$
✓
$\frac{1}{2}$
C
$1$
D
None of these

Answer

Correct option: B.

$\frac{1}{2}$

$\frac{1}{2}$

Solution:
Given:
The vertex and the focus of a parabola are V and S, respectively.
The given equation of parabola can be rewritten as follows:
$(y+3)^2-9+5+2 x=0$
$\Rightarrow(y+3)^2+2 x=4$
$\Rightarrow(y+3)^2=4-2 x$
$\Rightarrow(y+3)^2=-2(x-2)$
$\text { Let } Y=y+3, X=x-2$
Then, the equation of parabola becomes $Y^2=-2 X$.
$\text { Vertex }=(X=0, Y=0)=(x-2=0, y+3=0)=(x=2, y=-3)$
Comparing with $y^2=4 a x$ :
$4 \mathrm{a}=2 \Rightarrow \mathrm{a}=\frac{1}{2}$
$\text { Focus }=\left(\mathrm{X}=\frac{-1}{2}, \mathrm{Y}=0\right)=\left(\mathrm{x}-2=\frac{-1}{2}, \mathrm{y}+3=0\right)=\left(\mathrm{x}=\frac{3}{2}, \mathrm{y}=-3\right)$
$\Rightarrow \mathrm{SV}=\sqrt{\left(2-\frac{3}{2}\right)^2+(-3+3)^2}=\frac{1}{2}$

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MCQ 291 Mark

The length of latus rectum of the parabola $y^2+8 x-2 y+17=0$ is:

A
2
B
4
✓
8
D
16

Answer

Correct option: C.

Solution:
The given parabola is, $y^2+8 x-2 y+17=0$
$\Rightarrow\left(y^2-2 y+1\right)=-8 x-17+1=-8 x-16$
$\Rightarrow(y-1)^2=-8(x+2)$
Comparing with standard parabola $Y^2=-4 a X$
$Y=y-1, X=x+2, a=2$
Hence length of latus rectum is $=4 a=4 \times 2=8$

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MCQ 301 Mark

The coordinates of the focus of the parabola $y^2-x-2 y+2=0$ are

✓
$\Big(\frac{5}{4}, 1\Big)$
B
$\Big(\frac{1}{4}, 0\Big)$
C
$(1, 1)$
D
None of these

Answer

Correct option: A.

$\Big(\frac{5}{4}, 1\Big)$

$\Big(\frac{5}{4}, 1\Big)$

Solution:
Given:
The equation of the parabola is $y^2-x-2 y+2=0$.
$\Rightarrow(\mathrm{y}-1)-1=(\mathrm{x}-2)$
$(\mathrm{y}-1)=\mathrm{x}-1$
$\text { Let } \mathrm{X}=\mathrm{x}-1, \mathrm{Y}=\mathrm{y}-1$
$\mathrm{Y}=\mathrm{X}$
Comparing with $\mathrm{Y}=4 \mathrm{aX}$ :
$\mathrm{a}=\frac{1}{4}$
Focus =
$(X=a, Y=0)=\left(X=\frac{1}{4}, Y=0\right)=\left(x=\frac{1}{4}+1, y=1\right)=\left(x=\frac{5}{4}, y=1\right)$
Hence, the focus is at $\left(\frac{5}{4}, 1\right)$

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MCQ 311 Mark

If the circle $x^2 + y^2 = 9$ passesthrough (2, c) then c is equal to:

✓
$\sqrt{5}$
B
$\sqrt{6}$
C
$\sqrt{3}$
D
$\sqrt{7}$

Answer

Correct option: A.

$\sqrt{5}$

$\sqrt{5}$

Solution:
The equation of circle $x^2 + y^2 = 9$ The point is (2, c)
$\Rightarrow 2^2 + c^2= 9$
$4 + c^2 = 9$
$c^2 = 9 - 4$
$c^2 = 5$
$\text{c}=\sqrt{5}$

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MCQ 321 Mark

The number of tangents that can be drawn from $(1,2)$ to $x^2+y^2=5$ is:

A
0
✓
1
C
2
D
More than 2

Answer

Correct option: B.

Solution:
Given, point $(1,2)$ and equation of circle is $x^2+y^2=5$
Now, $x^2+y^2-5=0$
Put $(1,2)$ in this equation, we get
$1^2+2^2-5=1+4-5=5-5=0$
So, the point $(1,2)$ lies on the circle.
only one tangent can be drawn.

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MCQ 331 Mark

The focus of the parabola $y = 2x^2 + x$ is

A
$(0, 0)$
B
$\Big(\frac{1}{2}, \frac{1}{4}\Big)$
✓
$\Big(-\frac{1}{4},0\Big)$
D
$\Big(-\frac{1}{4}, \frac{1}{8}\Big)$

Answer

Correct option: C.

$\Big(-\frac{1}{4},0\Big)$

$\Big(-\frac{1}{4},0\Big)$

Solution:
Given:
Equation of the parabola = $y = 2x^2 + x$
$\Rightarrow\ \text{x}^2+\frac{\text{x}}{2}=\frac{\text{y}}{2}$
$\Rightarrow\ \Big(\text{x}+\frac{1}{4}\Big)^2=\frac{\text{y}}{2}+\frac{1}{16}$
$\Rightarrow\ \Big(\text{x}+\frac{1}{4}\Big)^2=\frac{8\text{y}+1}{16}$
$\Rightarrow\ \Big(\text{x}+\frac{1}{4}\Big)^2=\frac{1}{2}(\text{y}+\frac{1}{8})$
$\text{Let }\text{X}=\text{x}+\frac{1}{4},\text{Y}=\text{y}+\frac{1}{8}$
$\therefore\ \text{X}^2=\frac{1}{2}\text{Y}$
Comparing with X = 4aY
$\text{a}=\frac{1}{8}$
Focus $=(\text{X}=0,\ \text{Y}=\text{a})=\Big(\text{x}=\frac{-1}{4},\text{y}=0\Big)$
Hence, the focus is at $\Big(-\frac{1}{4},0\Big).$

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MCQ 341 Mark

The equation circle whose center is $(0,0)$ and radius is 4 is:

A
$x^2+y^2=4$
✓
$x^2+y^2=16$
C
$x^2+y^2=2$
D
None.

Answer

Correct option: B.

$x^2+y^2=16$

$x^2+y^2=16$

Solution:
The equation of circle is $x^2+y^2=r^2$
Here, the radius is 4 So the equation is $x^2+y^2=4^2$
$x^2+y^2=16$

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MCQ 351 Mark

Equation of the hyperbola whose vertices are $( \pm 3,0)$ and foci at $( \pm 5,0)$, is

✓
$16 x^2-9 y^2=144$
B
$9 x^2-16 y^2=144$
C
$25 x^2-9 y^2=225$
D
$9 x^2-25 y^2=81$

Answer

Correct option: A.

$16 x^2-9 y^2=144$

$16 x^2-9 y^2=144$

Solution:
The vertices of the hyperbola are $( \pm 3,0)$ and foci are $( \pm 5,0)$.
Thus, the value of a and ae are 3 and 5 , respectively.
Now, using the relation $b^2=a^2\left(e^2-1\right)$, we get:
$b^2=25-9$
$\Rightarrow b^2=16$
Equation of hyperbola is given below:
$\frac{x^2}{9}-\frac{y^2}{16}=1$
$16 x^2-9 y^2=144$

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MCQ 361 Mark

The vertex of the parabola $x^2+8 x+12 y+4=0$ is

✓
(-4, 1)
B
(4, -1)
C
(-4, -1)
D
(4, 1)

Answer

Correct option: A.

(-4, 1)

(-4, 1)

Solution:
Given:
$x^2+8 x+12 y+4=0$
$\Rightarrow(x+4)^2-16+12 y+4=0$
$\Rightarrow(x+4)^2+12 y-12=0$
$\Rightarrow(x+4)^2=-12(y-1)$
Let $\mathrm{X}=\mathrm{x}+4, \mathrm{Y}=\mathrm{y}-1$
$X^2=-12 Y$
Vertex $=(X=0, Y=0)=(x+4=0, y-1=0)=(x=-4, y=1)$
Hence, the vertex is at $(-4,1)$.

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MCQ 371 Mark

The equation of the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ will represent a real circle if:

A
$g^2 + f^2- c < 0$
✓
$\text{g}^{2} + \text{f}^{2} – \text{c} \underline{>} 0$
C
Always
D
None of these

Answer

Correct option: B.

$\text{g}^{2} + \text{f}^{2} – \text{c} \underline{>} 0$

$\text{g}^{2} + \text{f}^{2} – \text{c} \underline{>} 0$

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MCQ 381 Mark

For what value of $k$, does the equation $9 x^2+y^2=k\left(x^2-y^2-2 x\right)$ represents equation of a circle?

A
1
B
2
C
-1
✓
4

Answer

Correct option: D.

Solution:
$9 x^2-k x^2+y^2+k y^2+2 k x=0$
$x^2(9-k)+y^2(1+k)+2 k x=0$
for circle
$9-k=1+k$
So, $k=4$

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MCQ 391 Mark

A circle of radius 2 lies in the first quadrant and touches both the axes of co-ordinates. Then the equation of the circle with centre $(6,5)$ and touching the above circle externally is:

✓
$(x-6)^2+(y-5)^2=4$
B
$(x-6)^2+(y-5)^2=9$
C
$(x-6)^2+(y-5)^2=36$
D
None of these

Answer

Correct option: A.

$(x-6)^2+(y-5)^2=4$

$(x-6)^2+(y-5)^2=4$

Solution:
If $(h, k)$ is the center and the radius is $r$ then the equation of the circle is given by
$(x-h)^2+(y-k)^2=r^2$
Given that The center of the circle $(h, k)=(6,5)$ and the radius $\mathrm{r}=2$
$\therefore$ The equation of the circle is $(x-6)^2+(y-5)^2=4$

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MCQ 401 Mark

If the parabola $y^2= 4ax$ passes through the point $(3, 2)$, then the length of its latusrectum is:

A
$\frac{2}{3}$
✓
$\frac{4}{3}$
C
$\frac{1}{3}$
D
$4$

Answer

Correct option: B.

$\frac{4}{3}$

$\frac{4}{3}$

Solution:
Since, the parabola $y^2 = 4ax$ passes through the point $(3, 2)$
$\Rightarrow 2^2 = 4a \times 3$
$\Rightarrow 4 = 12a$
$\Rightarrow\text{a} =\frac{ 4}{12}$
$\Rightarrow\text{a}=\frac{1}{3}$
So, the length of latusrectum $= \text{4a} = 4 \times (\frac{1}{3}) = \frac{4}{3}$

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MCQ 411 Mark

If the circles $x^2 + y^2 = 9$ and $x^2 + y^2 + 8y + c = 0$ touch each other, then c is equal to:

✓
15
B
-15
C
16
D
-16

Answer

Correct option: A.

Solution:
The centre of the circle $x^2 + y^2 = 9$ is $(0, 0)$.
Let us denote it by $C_1$.
The centre of the circle $x^2 + y^2+ 8y + c = 0$ is $(0, -4)$.
Let us denote it by $C_2$.
The radius of $x^2 + y^2 = 9$ is $3$ units.
$x^2 + y^2+ 8y + c = 0$
$\Rightarrow(\text{x}-0)^2+(\text{y}+4)^2=16-\text{c}=(\sqrt{16-\text{c}})^2$
Therefore, the radius of the above circle is $\sqrt{16-\text{c}}$
Let the circles touch each other at P.
$\therefore\text{C}_1\text{C}_2=\text{PC}_2+\text{PC}_1$
$\Rightarrow\text{PC}_2=4-3=1$
$\Rightarrow\text{PC}_2-1=\sqrt{16-\text{c}}$
$\Rightarrow\text{c}=15$

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MCQ 421 Mark

Equation of the circle through origin which cuts intercepts of length a and b on axes is:

A
$x^2+y^2+a x+b y=0$
✓
$x^2+y^2-a x-b y=0$
C
$x^2+y^2+b x+a y=0$
D
None of these

Answer

Correct option: B.

$x^2+y^2-a x-b y=0$

$x^2+y^2-a x-b y=0$

Solution:
Centre of the circle is $\Big(\frac{\text{a}}{2},\ \frac{\text{b}}{2}\Big)$ and its radius is $\sqrt{\Big(\frac{\text{a}}{2}\Big)^2+\Big(\frac{\text{b}}{2}\Big)^2}=\frac{1}{2}\sqrt{\text{a}^2+\text{b}^2}$
Equation of circle:
$\Big(\text{x}-\frac{\text{a}}{2}\Big)^2+\Big(\text{y}-\frac{\text{b}}{2}\Big)^2=\frac{1}{4}(\text{a}^2+\text{b}^2)$
$\Rightarrow(2\text{x}-\text{a}^2)+(2\text{y}-\text{b})^2=(\text{a}^2+\text{b}^2)$
$\Rightarrow4\text{x}^2+\text{a}^2-4\text{ax}+4\text{y}^2+\text{b}^2-4\text{by}=\text{a}^2+\text{b}^2$
$\Rightarrow\text{x}^2-\text{ax}+\text{y}^2-\text{by}=0$

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MCQ 431 Mark

If $e_1$ and $e_2$ are respectively the eccentricities of the ellipse $\frac{\text{x}^2}{18}+\frac{\text{y}^2}{4}=1$ and the hyperbola $\frac{\text{x}^2}{9}-\frac{\text{y}^2}{4}=1,$ then the relation between $e_1$ and $e_2$ is

A
$3\text{e}_1^2 + \text{e}_2^2 = 2$
B
$\text{e}_1^2 + 2\text{e}_2^2 = 3$
✓
$2\text{e}_1^2 +\text{e}_2^2 = 3$
D
$\text{e}_1^2 + 3\text{e}_2^2 = 2$

Answer

Correct option: C.

$2\text{e}_1^2 +\text{e}_2^2 = 3$

$2\text{e}_1^2 +\text{e}_2^2 = 3$

Solution:
The standard from of the ellipse is $\frac{\text{x}^2}{18}+\frac{\text{y}^2}{4}=1,$ where $a^2 = 18$ and $b^2 = 4$.
So, the eccentricity is calculated in the following way:
$\text{b}2 = \text{a}2 (1 - \text{e}_1^2)$
$\Rightarrow4 = 18 (1 - \text{e}_1^2)$
$\Rightarrow\frac{2}{9}=1-\text{e}_1^2$
$\Rightarrow\text{e}_1^2=\frac{7}{9}$
The standard from of the hyperbola is $\frac{\text{x}^2}{9}-\frac{\text{y}^2}{4}=1,$ where $a^2 = 9$ and $b^2 = 4$.
So, the eccentricity is calculated in the following way:
$\text{b}^2 = \text{a}^2(\text{e}_2^2 - 1)$
$\Rightarrow4 = 9(\text{e}_2^2 - 1)$
$\Rightarrow\frac{4}{9}=\text{e}_2^2-1$
$\Rightarrow\text{e}_2^2=\frac{13}{9}$
$\therefore2\text{e}_1^2+\text{e}_2^2=\frac{2\times7}{9}+\frac{13}{9}$
$=\frac{27}{9}$
$=3$

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MCQ 441 Mark

If the focus of a parabola is (-2, 1) and the directrix has the equation x + y = 3, then its vertex is

A
$ (0, 3)$
B
$\Big(−1,\ \frac{1}{2}\Big)$
✓
$(−1, 2)$
D
$ (2, −1)$

Answer

Correct option: C.

$(−1, 2)$

Given:
The focus S is at (-2, 1) and the directrix is the line x + y - 3 = 0.
The slope of the line perpendicular to x + y - 3 = 0 is 1.
The axis of the parabola is perpendicular to the directrix and passes through the focus.
$\therefore$ Equation of the axis of the parabola = y - 1 = 1(x + 2) ...(1)
Intersection point of the directrix and the axis is the intersection point of (1) and x + y - 3 = 0.
Let the intersection point be K.
Therefore, the coordinates of K will be (0, 3).
Let (h, k) be the coordinates of the vertex, which is the mid-point of the segment joining K and the focus.
$\therefore\ \text{h}=\frac{0-2}{2},\ \text{k}=\frac{3+1}{2}$
h = -1, k = 2
Hence, the coordinates of the vertex are $(−1, 2).$

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MCQ 451 Mark

The centre of the circle $x^2+ y^2+ 10x - 20y + 100 = 0$ is:

A
(5, 10)
✓
(-5, 10)
C
(-5, -10)
D
(5, -10)

Answer

Correct option: B.

(-5, 10)

(-5, 10)

Solution:
Given the equation of the circle is $x^2+y^2+10 x-20 y+100=0$
or, $x^2+10 x+25+y^2-20 y+100=25$
or, $(x+5)^2+(y-10)^2=52$
From this equation it is clear that the centre is $(-5,10)$

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MCQ 461 Mark

The vertex of the parabola $(y-2)^2=16(x-1)$ is

✓
(1, 2)
B
(-1, 2)
C
(1, -2)
D
(2, 1)

Answer

Correct option: A.

(1, 2)

(1, 2)

Solution:
Given:
$(y-2)^2=16(x-1)$
$\text { Let } X=x-1, Y=y-2$
$\therefore Y^2=16 X$
$\text { Vertex }=(X=0, Y=0)=(x-1=0, y-2=0)=(x=1, y=2)$
Hence, the vertex is at $(1,2)$.

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MCQ 471 Mark

The number of integral values of $\lambda$ for which the equation $\text{x}^2+\text{y}^2+\lambda+(1-\lambda)\text{y}+5=0$ is the equation of a circle whose radius cannot exceed 5, is:

A
14
B
18
✓
16
D
None of these

Answer

Correct option: C.

$\sqrt{\Big(\frac{-\lambda}{2}^2\Big)+\Big(\frac{\lambda-1}{2}\Big)^2-5}\leq5$
$\Rightarrow\Big(\frac{-\lambda}{2}^2\Big)+\Big(\frac{\lambda-1}{2}\Big)\leq30$
$\lambda^2+(\lambda-1)^2\leq120$
$\Rightarrow2\lambda^2-2\lambda-199\leq0$
Using quadratic formula:
$\Rightarrow\lambda=\frac{2\pm\sqrt{2^2-4(2)(-119)}}{2(2)}$
$\Rightarrow\lambda=\frac{2\pm\sqrt{956}}{4}$
$\Rightarrow\lambda=\frac{1\pm\sqrt{239}}{2}$
$\Rightarrow\lambda=-7.23,\ 8.23$
$\Rightarrow-7.23\leq\lambda\leq8.23$
$\Rightarrow\lambda=-7,\ -6,\ -5,\ -4,\ -3,\ -2,\ -1\\0,\ 1,\ 2,\ 3,\ 4,\ 5,\ 6,\ 7,\ 8,\ $ $(\text{if}\ \lambda\in\text{Z})$
Thus, the number of integral values of $\lambda$ is 16.

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MCQ 481 Mark

If the line $\text{2x} - \text{y} + \lambda = 0$ is a diameter of the circle $x^2 + y^2 + 6x - 6y + 5 = 0$ then $\lambda=$

A
5
B
7
✓
9
D
11

Answer

Correct option: C.

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MCQ 491 Mark

In the parabola $y^2 = 4ax$, the length of the chord passing through the vertex and inclined to the axis at $\frac{\pi}{4}$ is

✓
$4\sqrt2\text{a}$
B
$2\sqrt2\text{a}$
C
$\sqrt2\text{a}$
D
None of these

Answer

Correct option: A.

$4\sqrt2\text{a}$

$4\sqrt2\text{a}$

Solution:

Let $O P$ be the chord.
Let the coordinates of $P$ be $\left(\mathrm{x}_1, \mathrm{y}_1\right)$.
From the figure, we have:
$O P^2=x_1^2+y_1^2 \ldots(1)$
And, $\tan \frac{\pi}{4}=\frac{y_1}{x_1}$
$\Rightarrow x_1=y_1 \ldots(2)$
Also, $\left(x_1, y_1\right)$ lies on the parabola.
$\therefore y_1{ }^2=4 \mathrm{ax}_{1 \ldots(3)}$
Using (2) and (3):
$x_1^2=4 a x_1 \Rightarrow x_1=4 a$
From (4), (1) and (2), we have:
$\mathrm{OP}^2=(4 a)^2+(4 a)^2=32 a^2$
$\Rightarrow \mathrm{OP}=4 \sqrt{2} \mathrm{a}$
Therefore, the length of the chord is $4 \sqrt{2} a$ a units.

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MCQ 501 Mark

If the coordinates of the vertex and the focus of a parabola are (-1, 1) and (2, 3) respectively, then the equation of its directrix is

✓
3x + 2y + 14 = 0
B
3x + 2y - 25 = 0
C
2x - 3y + 10 = 0
D
None of these

Answer

Correct option: A.

3x + 2y + 14 = 0

Given:
The vertex and the focus of a parabola are (-1, 1) and (2, 3), respectively.
$\therefore$ Slope of the axis of the parabola $=\frac{3-1}{2+1}=\frac{2}{3}$
Slope of the directrix $=\ \frac{-3}{2}$
Let the directrix intersect the axis at K (r, s).
$\therefore\ \frac{\text{r+2}}{2}=-1,\ \frac{\text{s}+3}{2}=1$
$\Rightarrow\ \text{r}=-4,\ \text{s}=-1$
Equation of the directrix:
$(\text{y}+1)=\frac{-3}{2}(\text{x}+4)$
$\Rightarrow\ 3\text{x}+2\text{y}+14=0$

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MCQ 511 Mark

Choose the correct answer. The equation of the ellipse whose focus is (1, -1), the directrix the line x - y - 3 = 0 and eccentricity $\frac{1}{2}$ is:

✓
$7 x^2+2 x y+7 y^2-10 x+10 y+7=0$
B
$7 x^2+2 x y+7 y^2+7=0$
C
$7 x^2+2 x y+7 y^2+10 x-10 y-7=0$
D
none

Answer

Correct option: A.

$7 x^2+2 x y+7 y^2-10 x+10 y+7=0$

$7 x^2+2 x y+7 y^2-10 x+10 y+7=0$

Solution:
Given that, foums of the ellipse is $S(1,-1)$ and the equation of directrix is $x-y-3=0$
Also, $e=\frac{1}{2}$
From definition of ellipse, for any point $P(x, y)$ on the ellipse, we have $S P=e P M$, where $M$ is foot of the perpendicular from point $P$ to the directrix.
$\therefore \sqrt{(x-1)^2+(y+1)^2}=\frac{1}{2} \frac{|x-y-3|}{\sqrt{2}}$
$\Rightarrow 8 x^2-16 x+16+8 y^2+16 y=x^2+y^2+9-2 x y+6 y-6 x$
$\Rightarrow 7 x^2+2 x y+7 y^2-10 x+10 y+7=0$

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MCQ 521 Mark

The eccentricity of the hyperbola $x^2 - 4y^2 = 1$

A
$\frac{\sqrt3}{2}$
✓
${\frac{\sqrt5}{2}}$
C
${\frac{2}{\sqrt3}}$
D
$\frac{2}{\sqrt5}$

Answer

Correct option: B.

${\frac{\sqrt5}{2}}$

${\frac{\sqrt5}{2}}$

Solution:
The equation of the hyperbola is $x^2 - 4y^2 = 1$.
This can be rewritten in the following way:
$\frac{\text{x}^2}{1}-\frac{\text{y}^2}{\frac{1}{4}}=1$
This is the standard form of a hyperbola, where a = 1 and $\text{b}^2=\frac{1}{4}.$
The value of eccentricity is calculated in the following way:
$\text{b}^2=\text{a}^2(\text{e}^2-1)$
$\Rightarrow\frac{1}{4}=(\text{e}^2-1)$
$\Rightarrow\text{e}^2=\frac{5}{4}$
$\Rightarrow\text{e}=\frac{\sqrt5}{4}$

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MCQ 531 Mark

The equations of the tangents to the ellipse $9\text{x}^2+16\text{y}^2=144$ from the point (2, 3) are:

A
y = 3, x = 5
B
x = 2, y = 3
C
x = 3, y = 2
✓
x + y = 5, y = 3

Answer

Correct option: D.

x + y = 5, y = 3

$9\text{x}^2+16\text{y}^2=144$
$\Rightarrow\frac{\text{x}^2}{16}+\frac{\text{y}^2}{9}=1$
Equation of the tangent in case of an ellipse is given by
$\text{y}=\text{mx}+\sqrt{\text{a}^2\text{m}^2+\text{b}^2}$
$\Rightarrow\text{y}=\text{mx}+\sqrt{16\text{m}^2+9}\ \dots(1)$
Substituting x = 2 and y = 3, we get:
$3=2\text{m}\pm\sqrt{16\text{m}^2+9}$
$\Rightarrow3-2\text{m}=\sqrt{16\text{m}^2+9}$
On squaring both sides, we get:
$(3-2\text{m})^2=(16\text{m}^2+9)$
$\Rightarrow9+4\text{m}^2-12\text{m}=(16\text{m}^2+9)$
$\Rightarrow12\text{m}^2+12\text{m}=0$
$\Rightarrow12\text{m}(\text{m+1})=0$
$\Rightarrow\text{m}=0,-1$
Substituting values of m in eq. (1), we get:
For $\text{m}=0,\ \text{y}=3$
For $\text{m}=-1,\ \text{y}=-\text{x}+5$ or $\text{x}+\text{y}=5$

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MCQ 541 Mark

The circle with radius 1 and centre being foot of the perpendicular from (5, 4) on y-axis, is:

A
$x^2+y^2-8 x-15=0$
B
$x^2+y^2-10 x+24=0$
✓
$x^2+y^2-8 y+15=0$
D
$x^2+y^2+2 y=0$

Answer

Correct option: C.

$x^2+y^2-8 y+15=0$

$x^2+y^2-8 y+15=0$

Solution:
Foot of perpendicular of $(5,4)$ on $y$-axis is $(0,4)$
$\therefore$ The equation of circle with
radius 1 cm is $(x-0)^2+(y-4)^2=1$
$\Rightarrow x^2+y^2-8 y+16$
$\Rightarrow x^2+y^2-8 y+15=0$

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MCQ 551 Mark

Choose the correct answer. The area of the circle centred at (1, 2) and passing through (4, 6) is:

A
$5\pi$
B
$10\pi$
✓
$25\pi$
D
none of these.

Answer

Correct option: C.

$25\pi$

Given that the centre of the circle is (1, 2)
Radius of the circle $=\sqrt{(4-1)^2+(6-2)^2}$
$=\sqrt{9+16}=5$
So, the area of the circle $=\pi\text{r}^2$
$=\pi\times(5)^2=25\pi$

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MCQ 561 Mark

Determine the area enclosed by the curve $x^2 - 10x + 4y + y^2 = 196$:

A
$15\pi$
✓
$225\pi$
C
$20\pi$
D
$17\pi$

Answer

Correct option: B.

$225\pi$

$225\pi$

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MCQ 571 Mark

Equation of the diameter of the circle $x^2 + y^2 − 2x + 4y = 0$ which passes through the origin is:

A
x + 2y = 0
B
x − 2y = 0
✓
2x + y = 0
D
2x − y = 0

Answer

Correct option: C.

2x + y = 0

2x + y = 0

Solution:
Let the diameter of the circle be y = mx.
Since the diameter of the circle passes through its centre, (1, -2) satisfies the equation of the diameter.
$\therefore$ m = -2
Substituting the value of m in the equation of diameter:
y = -2x
⇒ 2x + y = 0
Hence, the required equation of the diameter is 2x + y = 0.

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MCQ 581 Mark

The length of the transverse axis is the distance between the:

✓
Two vertices
B
Two Foci
C
Vertex and the origin
D
Focus and the vertex

Answer

Correct option: A.

Two vertices

The length of the transverse axis is the distance between two vertices.

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MCQ 591 Mark

The order of the differential equation of the family of parabolas whose length of latus rectum is fixed and axis is the x-axis:

✓
2
B
1
C
3
D
4

Answer

Correct option: A.

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MCQ 601 Mark

If $2\text{x}^2+\lambda\text{xy}+2\text{y}^2(\lambda-4)\text{x}+6\text{y}-5=0$ is the equation of a circle, then its radius is:

A
$3\sqrt{2}$
B
$2\sqrt{3}$
C
$2\sqrt{2}$
✓
None of these

Answer

Correct option: D.

None of these

The given equation is $2\text{x}^2+\lambda\text{xy}+2\text{y}^2+(\lambda-4)\text{x}+6\text{y}-5=0$ which can be rewritten as
$\text{x}^2+\frac{\lambda\text{xy}}{2}+\text{y}^2+\frac{(\lambda-4)}{2}\text{x}+3\text{y}-\frac{5}{2}=0.$
Comparing the given equation $\text{x}^2+\text{y}62+2\text{gx}+2\text{fy}+\text{c}=0$ with we get: $\lambda=0$
$\therefore\text{x}^2+\text{y}^2-2\text{x}+3\text{y}-\frac{5}{2}=0$
$\therefore$ Radius $=\sqrt{(-1)^2+\Big(\frac{3}{2}\Big)^2+\frac{5}{2}}=\sqrt{1+\frac{9}{4}+\frac{5}{2}}=\sqrt{\frac{23}{4}}=\frac{\sqrt{23}}{2}$

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MCQ 611 Mark

If the circle $x^2 + y^2 + 2ax + 8y + 16 = 0$ touches x-axis, then the value of a is:

A
$\pm16$
✓
$\pm4$
C
$\pm8$
D
$\pm1$

Answer

Correct option: B.

$\pm4$

$\pm4$

Solution:
The equation of the circle is $x^2 + y^2 + 2ax + 8y + 16 = 0$.
Its centre is (-a, -4) and its radius is a units.
Since the circle touches the x-axis, we have:
$\sqrt{(-\text{a}+\text{a})^2+(4-0)^2}=\text{a}$
$\Rightarrow\text{a}=\pm4$

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MCQ 621 Mark

The radius of the circle represented by the equation $3\text{x}^2+3\text{y}^2+(\lambda-6)\text{y}+3=0$ is:

✓
$\frac{3}{2}$
B
$\frac{\sqrt{17}}{2}$
C
$\frac{2}{3}$
D
None of these

Answer

Correct option: A.

$\frac{3}{2}$

The equation of the circle is $3\text{x}^2+3\text{y}^2+(\lambda-6)\text{y}+3=0$
$\therefore$ Coefficient of $\text{xy}=0$
$\Rightarrow\lambda=0$
$\therefore3\text{x}^2+3\text{y}^2+9\text{x}-6\text{y}+3=0$
$\Rightarrow\text{x}^2+\text{y}^2+3\text{x}-2\text{y}+1=0$
Therefore, the radius of the circle is $\sqrt{\Big(\frac{3}{2}\Big)^2+(-1)^2-1}=\frac{3}{2}.$

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MCQ 631 Mark

The equation of a hyperbola with foci on the $x-$axis is:

A
$\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2} = 1$
✓
$\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2} = 1$
C
$\text{x}^2 + \text{y}^2 = (\text{a}^2 + \text{b}^2)$
D
$\text{x}^2 - \text{y}^2 = (\text{a}^2 + \text{b}^2)$

Answer

Correct option: B.

$\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2} = 1$

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MCQ 641 Mark

The equation $x^2+ y^2 - 2x + 4y + 5 = 0$ represents:

✓
A point
B
A pair of straight lines
C
A circle of non zero radius
D
None of these

Answer

Correct option: A.

A point

A point

Solution:
$x^2+y^2-2 x+4 y+5=0$
$(x-1)^2+(y+2)^2-5+5=0$
$\Rightarrow(x-1)^2+(y+2)^2=0$
Since, radius is 0, its a point
Alternative method:
Here, a = b = 1
$\text{r}=\sqrt{1+4-5=0}$
a circle of radius 0. So, its a point.

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MCQ 651 Mark

The equation to the circle with centre (2, 1) and touches the line 3x + 4y - 5 is:

A
$x^2+y^2-4 x-2 y+5=0$
B
$x^2+y^2-4 x-2 y-5=0$
✓
$x^2+y^2-4 x-2 y+4=0$
D
$x^2+y^2-4 x-2 y-4=0$

Answer

Correct option: C.

$x^2+y^2-4 x-2 y+4=0$

$x^2+y^2-4 x-2 y+4=0$

Solution:
distance of pt. (2, 1) from line 3x + 4y - 5 is radius(r)
$\Rightarrow\text{r}=\frac{\mid6+4-5\mid}{5}=\frac{5}{5}=1$
⇒ Equation of circle is
⇒ $(x - 2)^2 + (y - 1)^2 = 1$
⇒ $x^2 + y^2 - 4x - 2y + 4 = 0$

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MCQ 661 Mark

The distance between the directrices of the hyperbola $\text{x}=8\sec\theta,\text{y}=8,$ is

✓
$8\sqrt2$
B
$16\sqrt2$
C
$4\sqrt2$
D
$6\sqrt2$

Answer

Correct option: A.

$8\sqrt2$

We have:
$\text{x}=8\sec\theta,\text{y}=8\tan\theta$
On squaring and subtracting:
$\text{x}^2-\text{y}^2=8\sec^2\theta-8\tan^2\theta$
$\Rightarrow\text{x}^2-\text{y}^2=8$
$\Rightarrow\frac{\text{x}^2}{8}-\frac{\text{y}^2}{8}=1$
$\therefore\text{a}=\text{b}=\text{c}$
Distance between the directrices of the hyperbola $=\frac{2\text{a}^2}{\sqrt{\text{a}^2+\text{b}^2}}$
Distance between the directrices $=\frac{2\times64}{\sqrt{64+64}}$
$=\frac{128}{8\sqrt2}$
$=\frac{16}{\sqrt2}$
$=8\sqrt2$

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MCQ 671 Mark

The equation of the conic with focus at (1, -1) directrix along x - y + 1 = 0 and eccentricity $\sqrt2$ is

✓
xy = 1
B
2xy + 4x - 4y - 1 = 0
C
$x^2 - y^2 = 1$
D
2xy - 4x + 4y + 1 = 0

Answer

Correct option: A.

xy = 1

2xy - 4x + 4y + 1 = 0

Solution:
Let P(x, y) be any point on the hyperbola.
Then, the distance of any point from the focus is eccentricity times the distance from the directrix.
$\therefore\sqrt{(\text{x}-1)^2+(\text{y}+1)^2}=\sqrt2\Big|\frac{\text{x}-\text{y}+1}{\sqrt2}\Big|$
Squaring both the sides, we get:
$(x - 1)^2 + (y + 1)^2 = (x - y + 1)^2$
$x^2 - 2x + 1 + y^2 + 1 + 2y = x^2 + y^2 + 1 - 2xy - 2y + 2x$
$2xy - 4x + 4y + 1 = 0$

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MCQ 681 Mark

The equation of a circle with radius $5$ and touching both the coordinate axes is:

A
$x^2+y^2 \pm 10 x \pm 10 y+5=0$
B
$x^2+y^2 \pm 10 x \pm 10 y=0$
✓
$x^2+y^2 \pm 10 x \pm 10 y+25=0$
D
$x^2+y^2 \pm 10 x \pm 10 y+51=0$

Answer

Correct option: C.

$x^2+y^2 \pm 10 x \pm 10 y+25=0$

$x^2+y^2 \pm 10 x \pm 10 y+25=0$

Solution:
Case I: If the circle lies in the first quadrant:
The equation of a circle that touches both the coordinate axes and hasradius a is $x^2+y^2-2 a x-2 a y+a^2=0$.
The given radius of the circle is 5 units, i.e. $a=5$.
Thus, the equation of the circle is $x^2+y^2-10 x-10 y+25=0$.
Case II: If the circle lies in the second quadrant:
The equation of a circle that touches both the coordinate axes and has radius $a$ is $x^2+y^2+2 a x-2 a y+a^2=0$.
The given radius of the circle is 5 units, i.e. $a=5$.
Thus, the equation of the circle is $x^2+y^2+10 x-10 y+25=0$.
Case III: If the circle lies in the third quadrant:
The equation of a circle that touches both the coordinate axes and has radius $a$ is $x^2+y^2+2 a x+2 a y+a^2=0$
The given radius of the circle is 5 units, i.e. $a=5$.
Thus, the equation of the circle is $x^2+y^2+10 x+10 y+25=0$.
Case IV: If the circle lies in the fourth quadrant:
The equation of a circle that touches both the coordinate axes and has radius $a$ is $x^2+y^2-2 a x+2 a y+a^2=0$.
The given radius of the circle is 5 units, i.e. $a=5$.
Thus, the equation of the circle is $x^2+y^2-10 x+10 y+25=0$.
Hence, the required equation of the circle is $x^2+y^2 \pm 10 x \pm 10 y+25=0$.

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MCQ 691 Mark

The perpendicular distance from the point (3, -4) to the line 3x - 4y + 10 = 0:

✓
7
B
8
C
9
D
10

Answer

Correct option: A.

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MCQ 701 Mark

If the point $(\lambda,\ \lambda+1)$ lies inside the region bounded by the curve $\text{x}=\sqrt{25-\text{y}^2}$ and y-axis, then $\lambda$ belongs to the interval:

✓
$(-1,\ 3)$
B
$(-4,\ 3)$
C
$(-\infty,\ -4)\cup(3,\ \infty)$
D
None of these

Answer

Correct option: A.

$(-1,\ 3)$

$(-1,\ 3)$

Solution:
The given equation of the curve is $x^2 + y^2 = 25$
Since $(\lambda,\ \lambda+1)$ lies inside the region bounded by the curve $x^2 + y^2 = 25$ and the y-axis, we have:
$\lambda^2+(\lambda+1)^2<25,$ provided $\lambda+1>0$
$\Rightarrow\lambda^2+\lambda^2+12\lambda<25,\ \lambda>-1$
$\Rightarrow2\lambda^2+2\lambda-24<0,\ \lambda>-1$
$\Rightarrow\lambda^2+\lambda-12<0,\ \lambda>-1$
$\Rightarrow(\lambda-3)(\lambda+4)<0,\ \lambda>-1$
$\Rightarrow-4<\lambda<3,\ \lambda>-1$
$\Rightarrow\lambda\in(-1,\ 3)$

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MCQ 711 Mark

The center of the circle $4 x^2+4 y^2-8 x+12 y-25=0$ is:

✓
(2, -3)
B
(-2, 3)
C
(-4, 6)
D
(4, -6)

Answer

Correct option: A.

(2, -3)

(2, -3)

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MCQ 721 Mark

The equation of circle center at (0, 0) and Radius 8cm:

✓
$x^2+y^2=64 cm$
B
$x^2+y^2=8$
C
$x^2+y^2=16$
D
$x^2+y^2=4$

Answer

Correct option: A.

$x^2+y^2=64 cm$

$x^2+y^2=64 cm$

Solution:
The equation of circle is $x^2+y^2=r^2$
$x^2+y^2=8^2$
$x^2+y^2=64$

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MCQ 731 Mark

The area of an equilateral triangle inscribed in the circle $x^2 + y^2 - 6x - 8y - 25 = 0$ is:

✓
$\frac{225\sqrt{3}}{6}$
B
$25\pi$
C
$50\pi-100$
D
None of these

Answer

Correct option: A.

$\frac{225\sqrt{3}}{6}$

$\frac{225\sqrt{3}}{6}$

Solution:

Let ABC be the required equilateral triangle.
The equation of the circle is $x^2 + y^2 - 6x - 8y - 25 = 0$.
Therefore, coordinates of the centre O is (3, 4).
Radius of the circle $=\text{OA}=\text{OB}=\text{OC}=\sqrt{9+16+25}=5\sqrt{2}$
In $\Delta\text{BOD},$ we have:
$\sin60^\circ=\frac{\text{DB}}{\text{BO}}$
$\Rightarrow\text{DB}=\frac{\sqrt{3}}{2}(5\sqrt{2})$
$\Rightarrow\text{BC}=2\text{BD}-\sqrt{3}\big(5\sqrt{2}\big)=5\sqrt{6}$
Now, area of $\triangle\text{ABC}=\frac{\sqrt{3}}{4}\text{BC}^2=\big(5\sqrt{6}\big)^2\\=\frac{\sqrt{3}(150)}{4}=\frac{\sqrt{3}(75)}{2}=\frac{\sqrt{3}(225)}{6}$ square units

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MCQ 741 Mark

The distance between the foci of a hyperbola is 16 and its eccentricity is $\sqrt2$ , then equation of the hyperbola is

A
$x^2+y^2=32$
B
$x^2-y^2=16$
C
$x^2+y^2=16$
✓
$x^2-y^2=32$

Answer

Correct option: D.

$x^2-y^2=32$

$x^2-y^2=32$

Solution:
The distance between the foci is 2ae.
$\therefore$ 2ae = 16
⇒ ae = 8
$\text{e}=\sqrt2$
$\therefore\text{a}\sqrt2=8$
$\Rightarrow\text{a}=4\sqrt2$
Also, $b^2 = a^2(e^2 − 1)$
$\Rightarrow b^2 = 32(2 − 1)$
$\Rightarrow b^2 = 32$
Standard form of the hyperbola is given below:
$\frac{\text{x}^2}{32}-\frac{\text{y}^2}{32}=1$
$\text{x}^2-\text{y}^2=32$

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MCQ 751 Mark

Which of the following points lie on the parabola $x^2=4 a y$ ?

A
$x=a t^2, y=2 a t$
B
$x=2 a t, y=a t^2$
C
$x=2 a t^2, y=a t$
✓
$x=2 a t, y=a t^2$

Answer

Correct option: D.

$x=2 a t, y=a t^2$

$x=2 a t, y=a t^2$

Solution:
Substituting $x=2 a t, y=a t^2$ in the given equation:
$(2 a t)^2=4 a\left(a t^2\right)$
$\Rightarrow 4 a^2 t^2=4 a^2 t^2$
Hence, (2at, at ${ }^2$ ) lies on the parabola $x^2=4 a y$.

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MCQ 761 Mark

The equation of the circle concentric with $x^2+y^2-3 x+4 y-c=0$ and passing through $(-1,-2)$ is:

A
$x^2+y^2-3 x+4 y-1=0$
✓
$x^2+y^2-3 x+4 y=0$
C
$x^2+y^2-3 x+4 y+2=0$
D
None of these

Answer

Correct option: B.

$x^2+y^2-3 x+4 y=0$

$x^2+y^2-3 x+4 y=0$

Solution:
The centre of the circle $x^2 + y^2 - 3x + 4y - c = 0$ is $\Big(\frac{3}{2},\ -2\Big).$
Therefore, the centre of the required circle is $\Big(\frac{3}{2},\ -2\Big).$
The equation of the circle is $\Big(\text{x}-\frac{3}{2}\Big)^2+(\text{y}+2)^2=\text{a}^2. \ ......(1)$
Also, circle (1) passes through (-1, -2).
$\therefore\Big(-1-\frac{3}{2}\Big)^2+\Big(-2+2\Big)^2=\text{a}^2$
$\Rightarrow\text{a}=\frac{5}{2}$
Substituting the value of a in equation (1):
$\Big(\text{x}-\frac{3}{2}\Big)^2+(\text{y}+2)^2=\Big(\frac{5}{2}\Big)^2$
$\Rightarrow\frac{(2\text{x}-3)^2}{4}+(\text{y}+2)^2=\frac{25}{4}$
$\Rightarrow(2\text{x}-3)^2+4(\text{y}+2)^2=25$
$\Rightarrow\text{x}^2+\text{y}^2-3\text{x}+4\text{y}=0$
Hence, the required equation of the circle is $x^2 + y^2 - 3x + 4y = 0$.

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MCQ 771 Mark

The equation of the circle passing through (2, 0) and (0, 4) and having the minimum radius is:

A
$x^2+y^2=20$
✓
$x^2+y^2-2 x-4 y=0$
C
$x^2+y^2=4$
D
$x^2+y^2=16$

Answer

Correct option: B.

$x^2+y^2-2 x-4 y=0$

$x^2+y^2-2 x-4 y=0$

Solution:
Given, points are (2, 0) and (0, 4)
$\therefore$ equation of circle is (x - 2) (x - 0) + (y - 0) (y - 4) = 0
By expanding, we get
$x^2 - 2x + y^2 - 4y = 0$

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MCQ 781 Mark

Choose the correct answer. The equation of a circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length 3a is:
[Hint: Centroid of the triangle coincides with the centre of the circle and the radius of the circle is $\frac{2}{3}$ of the length of the mediam]

A
$x^2+y^2=9 a^2$
B
$x^2+y^2=16 a^2$
✓
$x^2+y^2=4 a^2$
D
$x^2+y^2=a^2$

Answer

Correct option: C.

$x^2+y^2=4 a^2$

$x^2+y^2=4 a^2$

Solution:
Let ABC be an equilateral triangle in which mediam AD = 3a
Centre of the circle is same as the centroid of the triangle i.e., $(0, 0)$

AG : GD = 2 : 1
So, $\text{AG}=\frac{2}{3}\text{AD}=\frac{2}{3}\times3\text{a}=2\text{a}$
$\therefore$ The equation of the circle is,
$(x - 0)^2 + (y - 0)^2= (2a)^2$
$\Rightarrow x^2 + y^2 = 4a^2$

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MCQ 791 Mark

The equation of the circle passing through (3, 6) and whose centre is (2, -1) is:

✓
$x^2+y^2-4 x+2 y=45$
B
$x^2+y^2-4 x-2 y+45=0$
C
$x^2+y^2+4 x-2 y=45$
D
$x^2+y^2-4 x+2 y+45=0$

Answer

Correct option: A.

$x^2+y^2-4 x+2 y=45$

$x^2+y^2-4 x+2 y=45$

Solution:
Equation of circle, $(\text{x} - 2)^2 + (\text{y} -( -1))^2= \Big(\sqrt{{(3-2)^2+(6}-(-1))^2\Big)}^2$
$\text{x}^2 - 4\text{x} + 4 + \text{y}^2 + 2\text{y} + 1=(\sqrt{1+49})^2\therefore\text{x}^2+\text{y}^2-4\text{x}+2\text{y}=45$ Equation of circle.

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MCQ 801 Mark

The eccentricity of the ellipse $\frac{\text{x}^2}{\text{b}^2}+\frac{\text{y}^2}{\text{y}^2}=1$ if its latus rectum is equal to one half of its minor axis, is:

A
$\frac{1}{\sqrt{2}}$
✓
$\frac{\sqrt{3}}{2}$
C
$\frac{1}{2}$
D
$\text{none of these}$

Answer

Correct option: B.

$\frac{\sqrt{3}}{2}$

According to the question, the latus rectum is half its minor axis.
i.e. $\frac{2\text{b}^2}{\text{a}}=\frac{1}{2}\times2\text{b}$
$\Rightarrow2\text{b}^2=\text{ab}$
$\Rightarrow\text{a}=2\text{b}$
Now, $\text{e}\sqrt{1-\frac{\text{b}^2}{\text{a}^2}}$
$\Rightarrow\text{e}=\sqrt{1-\frac{\text{b}^2}{4\text{b}^2}}$
$\Rightarrow\text{e}=\sqrt{1-\frac{1}{4}}$
$\Rightarrow\text{e}=\frac{\sqrt{3}}{2}$

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MCQ 811 Mark

Assertion: If the equation of a circle is ($x + 1)^2+ (y - 1)^2 = 4$, then its radius is 4. Reason: Equation of a circle with radius r is given by, $(x -a)^2 + (y - b)^2 = r2$.

A
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
B
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
C
Assertion is correct but Reason is incorrect
✓
Assertion is incorrect but Reason is correct

Answer

Correct option: D.

Assertion is incorrect but Reason is correct

Assertion is incorrect but Reason is correct

Solution:
$(x + 1)^2 + (y - 1)^2= 2^2$ Radius = 2 Centre (-1, 1) Assertion is incorrect but reason is correct.

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MCQ 821 Mark

The equation of the circle which touches $x$-axis at $(0,0)$ and touches the line $3 x+4 y-5=0$ is:

A
$x^2+y^2-4 y=0$
B
$x^2+y^2-10 y=0$
C
$x^2+y^2+10 x=0$
✓
$x^2+y^2+10 y=0$

Answer

Correct option: D.

$x^2+y^2+10 y=0$

$x^2+y^2+10 y=0$

Solution:
Equation of circle touching $x$-axis at $(0,0)$, means centre of circle lie on $Y$-axis i.e. $(0, k)$.
$(x-0)^2+(y-k)^2=k^2$
$S: x^2+y^2-2 k y=0$.... (1)
Circle S touches $3 x+4 y-5=0$
$\therefore \mathbf{k}=\frac{4 \mathbf{k}-5}{5}$
$5 k=4 k-5$
$k=-5$
$\therefore$ Equation of circle is
$\Rightarrow x^2+y^2+10 y=0$

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MCQ 831 Mark

Choose the correct answer. Equation of a circle which passes through $(3,6)$ and touches the axes is:

A
$x^2+y^2+6 x+6 y+3=0$
B
$x^2+y^2-6 x-6 y-9=0$
✓
$x^2+y^2-6 x-6 y+9=0$
D
none of these.

Answer

Correct option: C.

$x^2+y^2-6 x-6 y+9=0$

$x^2+y^2-6 x-6 y+9=0$

Solution:
Given that the circle touches both axes.
Therefore, equation of the circle is, $(x - a)^2 + (y - a)^2 = a^2$
Circle passes through the point (3, 6)
$\therefore (3 - a)^2 + (6 - a)^2 = a^2$
$\Rightarrow a^2 - 18a + 45 = 0$
$\Rightarrow (a - 3)(a - 15) = 0$
$\therefore a = 3, a = 15$
For a = 3, the equation of circle is,
$(x - 3)^2 + (y - 3)^2 = 9$
$\Rightarrow x^2 + y^2 - 6x - 6y + 9 = 0$

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MCQ 841 Mark

If $(-3,2)$ lies on the circle $x^2+y^2+2 g x+2 f y+c=0$ which is concentric with the circle $x^2+y^2+6 x+8 y-5=0$, then $\mathrm{c}=$

A
11
✓
-11
C
24
D
None of these

Answer

Correct option: B.

-11

Solution:
The centre of the circle $x^2+y^2+6 x+8 y-5=0$ is $(-3,-4)$.
The circle $x^2+y^2+2 g x+2 f y+c=0$ is concentric with the circle $x^2+y^2+6 x+8 y-5=0$.
Thus, the centre of $x^2+y^2+2 g x+2 f y+c=0$ is $(-3,-4)$.
$\therefore g=3, f=4$
Also, it is given that $(-3,2)$ lies on the circle $x^2+y^2+2 g x+2 f y+c=0$.
$\therefore(-3)^2+2^2+2(3)(-3)+2(4)(2)+c=0$
$\Rightarrow 9+4-18+16+c=0$
$\Rightarrow c=-11$

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MCQ 851 Mark

Choose the correct answer. If the focus of a parabola is (0, -3) and its directrix is y = 3, then its equation is:

✓
$x^2=-12 y$
B
$x^2=12 y$
C
$y^2=-12 x$
D
$y^2=12 x$

Answer

Correct option: A.

$x^2=-12 y$

$x^2=-12 y$

Solution:
According to the definition of parabola,
$\sqrt{(\text{x}-0)^2+(\text{y}+3)^2}=\Bigg|\frac{\text{y}-3}{\sqrt{(0)^2+(1)^2}}\Bigg|$
$\Rightarrow\sqrt{\text{x}^2+\text{y}^2+9+6\text{y}}=|\text{y}-3|$
Squaring both sides, we get
$x^2+y^2+9+6 y=y^2+9-6 y$
$\Rightarrow x^2+9+6 y=9-6 y$
$\Rightarrow x^2=-12 y$

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MCQ 861 Mark

Choose the correct answer. If the vertex of the parabola is the point (-3, 0) and the directrix is the line x + 5 = 0, then its equation is:

✓
$y^2=8(x+3)$
B
$x^2=8(y+3)$
C
$y^2=-8(x+3)$
D
$y^2=8(x+5)$

Answer

Correct option: A.

$y^2=8(x+3)$

$y^2=8(x+3)$

Solution:
Given that vertex $\equiv(-3,0)$ and directrix, x + 5 = 0

So, focus $\equiv\text{S}(-1,0)$
For any point of parabola P(x, y) we have,
$\text{SP}=\text{PM}$
$\Rightarrow\sqrt{(\text{x}+1)+\text{y}^2}=|\text{x}+5|$
$\Rightarrow\text{x}^2+2\text{x}+1+\text{y}^2=\text{x}^2+10\text{x}+25$
$\Rightarrow\text{y}^2=8\text{x}+24$
$\Rightarrow\text{y}^2=8(\text{x}+3)$

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MCQ 871 Mark

What is the equation of a circle with center $(-3,1)$ and radius 7 :

A
$(x-3)^2+(y+1)^2=7$
B
$(x-3)^2+(y+1)^2=49$
C
$(x+3)^2+(y-1)^2=7$
✓
$(x+3)^2+(y-1)^2=49$

Answer

Correct option: D.

$(x+3)^2+(y-1)^2=49$

$(x+3)^2+(y-1)^2=49$

Solution:
The general equation of a circle with center at $(a, b)$ and radius $r$ is $(x-a)^2+(y-b)^2=r^2$
So substituting the values we get the equation of the circle is $(x+3)^2+(y-1)^2=7^2=49$

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MCQ 881 Mark

The eccentricity of the hyperbola whose latus-rectum is half of its transverse axis, is

A
$\frac{1}{\sqrt2}$
B
$\sqrt{\frac{2}{3}}$
✓
$\sqrt{\frac{3}{2}}$
D
None of these.

Answer

Correct option: C.

$\sqrt{\frac{3}{2}}$

The lengths of the latus rectum and the transverse axis are $\frac{2\text{b}^2}{\text{a}}\text{ and }2\text{a},$ respectively.
According to the given statement, length of the latus rectum is half of its transverse axis.
$\therefore\frac{2\text{b}^2}{\text{a}}=\frac{1}{2}\times2\text{a}$
$\Rightarrow\frac{2\text{b}^2}{\text{a}}=\text{a}$
$\Rightarrow2\text{b}^2=\text{a}$
Eccentricity, $\text{e}=\frac{\sqrt{\text{a}^2+\text{b}^2}}{\text{a}}$
Substituting the value $\text{b}^2=\frac{\text{a}^2}{2},$ we get:
$\text{e}=\frac{\sqrt{\text{a}^+\frac{\text{a}}{2}}}{\text{a}}$
$=\frac{\text{a}\sqrt{\frac{3}{2}}}{\text{a}}$
$=\sqrt{\frac{3}{2}}$
$\therefore$ Eccentricity is $\sqrt{\frac{3}{2}}$

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MCQ 891 Mark

If $e_1$ is the eccentricity of the conic $9 x^2+4 y^2=36$ and $e_2$ is the eccentricity of the conic $9 x^2-4 y^2=36$, then

A
$\text{e}_1^2-\text{e}_2^2=2$
✓
$2<\text{e}_2^2-\text{e}_1^2<3$
C
$\text{e}_2^2-\text{e}_1^2=2$
D
$\text{e}_2^2-\text{e}_1^2>3$

Answer

Correct option: B.

$2<\text{e}_2^2-\text{e}_1^2<3$

$2<\text{e}_2^2-\text{e}_1^2<3$

Solution:
The conic $9x^2 + 4y^2 = 36$ can rewritten in the following way:
$\frac{9\text{x}^2}{36}+\frac{4\text{y}^2}{36}=1$
$\Rightarrow\frac{\text{x}^2}{4}+\frac{\text{y}^2}{9}=1$
This is the standard equation of an ellipse.
$\therefore b^2 = a^2(1−e_1)^2$
$\Rightarrow9=4(1-\text{e}_1)^2$
$\Rightarrow(\text{e}_1)^2=\frac{-5}{4}$
The conic $9x^2 − 4y^2 = 36$ can rewritten in the following way:
$\frac{9\text{x}^2}{36}-\frac{4\text{y}^2}{36}=1$
$\Rightarrow\frac{\text{x}^2}{4}-\frac{\text{y}^2}{9}=1$
This is the standard equation of a hyperbola.
$\therefore b^2 = a^2(e_2^2 − 1)$
$\Rightarrow9=4(\text{e}_2^2-1)$
$\Rightarrow(\text{e}_2)^2=\frac{13}{4}$
$\therefore\text{e}_2^2-\text{e}_1^2=\frac{13}{4}+\frac{5}{4}=2.5$

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MCQ 901 Mark

If the circles $x^2 + y^2 + 2ax + c = 0$ and $x^2 + y^2 + 2by + c = 0$ touch each other, then:

✓
$\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}=\frac{1}{\text{c}}$
B
$\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}=\frac{1}{\text{c}}$
C
$\text{a}+\text{b}=2\text{c}$
D
$\frac{1}{\text{a}}+\frac{1}{\text{b}}=\frac{2}{\text{c}}$

Answer

Correct option: A.

$\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}=\frac{1}{\text{c}}$

$\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}=\frac{1}{\text{c}}$

Solution:
Given:
$x^2 + y^2 + 2ax + c = 0$ ....... (1)
And, $x^2 + y^2 + 2by + c = 0$ ........ (2)
For circle (1), we have:
Centre = $(-a, 0) = C_1$
For circle (2), we have:
Centre = $(0,-b) = C_2$
Let the circles intersect at point P.
$\therefore$ Coordinates of P = Mid point of $C_1C_2$
$\Rightarrow$ Coordinates of P $=\Big(\frac{-\text{a}+0}{2},\ \frac{0-\text{b}}{2}\Big)=\Big(\frac{-\text{a}}{2},\ \frac{-\text{b}}{2}\Big)$
Now, we have:
$PC_1$ = radius of (1)
$\Rightarrow\sqrt{(-\text{a}+\frac{\text{a}}{2})^2}+\Big(0-\frac{\text{b}}{2}\Big)^2=\sqrt{\text{a}^2-\text{c}}$
$\Rightarrow\frac{\text{a}^2}{4}+\frac{\text{b}}{4}^2=\text{a}^2-\text{c}\ .....(3)$
Also, radius of circle (1) = radius of circle (2)
$\Rightarrow\sqrt{\text{a}^2-\text{c}}=\sqrt{\text{b}^2-\text{c}}$
$\Rightarrow\text{a}^2=\text{b}^2\ .....(4)$
From (3) and (4), we have:
$\frac{\text{a}^2}{2}=\text{a}^2-\text{c}$
$\Rightarrow\frac{\text{a}^2}{2}=\text{c}$
$\Rightarrow\frac{2}{\text{a}^2}=\frac{1}{\text{c}}$
$\Rightarrow\frac{1}{\text{a}^2}+\frac{1}{\text{a}^2}=\frac{1}{\text{c}}$
$\Rightarrow\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}=\frac{1}{\text{c}}$

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MCQ 911 Mark

For the ellipse $12\text{x}^2+4\text{y}^2+24\text{x}-16\text{y}+25=0$

A
centre is (-1, 2)
B
lengths of the axes are $\sqrt{3}$ and 1
C
eccentricity $=\sqrt{\frac{2}{3}}$
✓
all of these.

Answer

Correct option: D.

all of these.

Disclaimer : The equation should be $12\text{x}^2+4\text{y}^2+24\text{x}-16\text{y}+24=0$ instead of $12\text{x}^2+4\text{y}^2+24\text{x}-16\text{y}+25=0.$
$12\text{x}^2+4\text{y}^2+24\text{x}-16\text{y}+24=0$
$\Rightarrow12\big(\text{x}^2+2\text{x}\big)+4\big(\text{y}^2-4\text{y}\big)=-24$
$\Rightarrow12\big(\text{x}^2+2\text{x}+1\big)+4\big(\text{y}^2-4\text{y}+4\big)=-24+12+16$
$\Rightarrow12\big(\text{x}+1\big)^2+4\big(\text{y}-2\big)^2=4$
$\Rightarrow\frac{(\text{x}+1)^2}{3}+\frac{(\text{y}-2)^2}{1}=1$
So, the centre is a $(-1,\ 2).$
Here, $\text{a}=\sqrt{3}$ and $\text{b}=1$
The lengths of the axes are $\sqrt{3}$ and 1.
Now, $\text{e}=\sqrt{1-\frac{\text{b}62}{\text{a}^2}}$
$\text{e}=\sqrt{1-\frac{1}{3}}$
$\Rightarrow\text{e}=\sqrt{\frac{2}{3}}$

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MCQ 921 Mark

The parametric equations of a parabola are $x=t^2+1, y=2 t+1$. The cartesian equation of its directrix is

✓
x = 0
B
x + 1 = 0
C
y = 0
D
None of these

Answer

Correct option: A.

x = 0

x = 0

Solution:
Given:
$x=t^2+1$
$y=2 t+1$
From (1) and (2):
$x=\left(\frac{y-1}{2}\right)^2+1$
On simplifying:
$(y-1)^2=4(x-1)$
Let $\mathrm{Y}=\mathrm{y}-1$ and $\mathrm{X}=\mathrm{x}-1$
$\therefore Y^2=4 X$
Comparing it with $y^2=4 a x:$
$a=1$
Therefore, the equation of the directrix is $X=-a$, i.e. $x-1=-1 \Rightarrow x=0$

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MCQ 931 Mark

Choose the correct answer. If the parabola $y^2 = 4ax$ passes through the point (3, 2), then the length of its latus rectum is:

A
$\frac{2}{3}$
✓
$\frac{4}{3}$
C
$\frac{1}{3}$
D
$4$

Answer

Correct option: B.

$\frac{4}{3}$

$\frac{4}{3}$

Solution:
Given parabola is $y^2 = 4ax$
If the parabola is passing through (3, 2)
Then $(2)^2 = 4a \times 3$
$\Rightarrow 4 = 12a$
$\Rightarrow\text{a}=\frac{1}{3}$
Nowm length of the latus rectum $=4\text{a}=4\times\frac{1}{3}=\frac{4}{3}$

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MCQ 941 Mark

The focus of the parabola $y^2 = 8x$ is:

A
(0, 2)
✓
(2, 0)
C
(0, -2)
D
(-2, 0)

Answer

Correct option: B.

(2, 0)

(2, 0)

Solution:
Given parabola equation $y^2 = 8x …(1)$
Here, the coefficient of x is positive and the standard form of parabola is $y^2 = 4ax … (2)$
Comparing (1) and (2), we get
4a = 8
$\text{a} = \frac{8}{4} = 2$
We know that the focus of parabolic equation $y^2 = 4ax$ is $(a, 0)$.
$\therefore$ The focus of the parabola $y^2 = 8x$ is $(2, 0)$.

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MCQ 951 Mark

The equation of the directrix of the parabola whose vertex and focus are (1, 4) and (2, 6) respectively is

✓
x + 2y = 4
B
x - y = 3
C
2x + y = 5
D
x + 3y = 8

Answer

Correct option: A.

x + 2y = 4

Given:
The vertex and the focus of a parabola are (1, 4) and (2, 6), respectively.
$\therefore$ Slope of the axis of the parabola $= \frac{6-4}{2-1}=2$
Slope of the directrix $=\ \frac{-1}{2}$
Let the directrix intersect the axis at K (r, s).
$\therefore\ \frac{\text{r}+2}{2}=1,\ \frac{\text{s}+6}{2}=4$
$\Rightarrow\ \text{r}=0,\ \text{s}=2$
Equation of the directrix:
$(\text{y}-2)=\frac{-1}{2}(\text{x}-0)$
⇒ x + 2y = 4

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MCQ 961 Mark

Equation of the parabola having focus (3, 2) and Vertex (-1, 2) is:

A
$(x+1)^2=16(y-2)$
B
$(x-1)^2=16(y+2)$
✓
$(y-2)^2=16(x+1)$
D
$(y+2)^2=16(x-1)$

Answer

Correct option: C.

$(y-2)^2=16(x+1)$

$(y-2)^2=16(x+1)$

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MCQ 971 Mark

Latus rectum of a parabola is a $........$ line segment with respect to the axis of the parabola through the focus whose endpoints lie on the parabola:

✓
Perpendicular
B
Parallel
C
Tilted
D
None of these

Answer

Correct option: A.

Perpendicular

Consider the above image which shows that the latus rectum of a parabola is a line segment perpendicular to the axis of the parabola, through the focus and whose end points lie on the parabola.

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MCQ 981 Mark

If the eccentricity of the hyperbola $x^2-y^2 \sec ^2 \alpha=5$ is $\sqrt{3}$ times the eccentricity of the ellipse $x^2 \sec ^2 \alpha+y^2=25$, then $\alpha=$

A
$\frac{\pi}{6}$
✓
$\frac{\pi}{4}$
C
$\frac{\pi}{3}$
D
$\frac{\pi}{2}$

Answer

Correct option: B.

$\frac{\pi}{4}$

$\frac{\pi}{4}$

Solution:
The hyperbola $\text{x}^2 − \text{y}^2 \sec^2\alpha = 5$ can be rewritten in the following way:
$\frac{\text{x}^2}{5}-\frac{\text{y}^2}{5\cos^2\text{a}}=1$
This is the standard form of a hyperbola, where $ \text{a}^2 = 5 \text{ and } \text{b}^2 = 5\cos^2\alpha.$
$\Rightarrow\text{b}^2 = \text{a}^2(\text{e}_1^2 − 1)$
$⇒ 5\cos^2\alpha=5(\text{e}_1^2−1)$
$\Rightarrow\text{e}_1^2=\cos^2\alpha+1...(1)$
The ellipse $\text{x}^2\sec^2\alpha+\text{y}^2=25$ can be rewritten in the following way:
$\frac{\text{x}^2}{25\cos^2\alpha}+\frac{\text{y}^2}{25}=1$
This is the standard form of an ellipse, where $\text{a}^2=25\text{ and }\text{b}^2=25\cos^2\alpha$
$\text{b}^2=\text{a}^2(1-\text{e}_2^2)$
$\Rightarrow\text{e}_2^2=1-\cos^2\alpha$
$\Rightarrow\text{e}_2^2=\sin^2\alpha...(2)$
According to the question,
$\cos^2\alpha+1=3(\sin^2\alpha)$
$\Rightarrow2=4\sin^2\alpha$
$\Rightarrow\sin\alpha=\frac{1}{\sqrt2}$
$\Rightarrow\alpha=\frac{\pi}{4}$

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MCQ 991 Mark

Find the equation to the circle which touches the axis of $y$ at the origin and passes through the point (b, c):

A
$b x^2+b y^2-\left(b^2+c^2\right) y=0$
✓
$b x^2+b y^2-\left(b^2+c^2\right) x=0$
C
$b x^2+b y^2+\left(b^2+c^2\right) y=-1$
D
$b x^2+c y^2-\left(b^2+c^2\right) x=1$

Answer

Correct option: B.

$b x^2+b y^2-\left(b^2+c^2\right) x=0$

$b x^2+b y^2-\left(b^2+c^2\right) x=0$

Solution:
Equation of circle which touches the $y$-axis at origin is $x^2+y^2+2 g x+d=0$
Since the circle passes through origin, $d=0$ Thus the equation becomes, $x^2+y^2+2 g x=0 \ldots .$. (1)
The equation passes through $(b, c)$
so, $b^2+c^2+2 g b=0$
$\therefore \mathrm{g}=\frac{-\mathrm{b}^2-\mathrm{c}^2}{2 \mathrm{~b}}$
So, putting the value of $g$ in (1) we getb $x^2+b y^2-\left(b^2+c^2\right) x=0$

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MCQ 1001 Mark

The equation of parabola with vertex at origin and directrix $x-2=0$ is:

A
$y^2=-4 x$
B
$y^2=4 x$
✓
$y^2=-8 x$
D
$y^2=8 x$

Answer

Correct option: C.

$y^2=-8 x$

$y^2=-8 x$

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MCQ 1011 Mark

The number of tangents that can be drawn from $(1,2)$ to $x^2+y^2=5$ is:

A
0
✓
1
C
2
D
more than 2

Answer

Correct option: B.

Solution:
Given circle equation: $x^2+y^2=5$
$x^2+y^2-5=0 \ldots$
Now, substitute $(1,2)$ in equation (1), we get
Circle Equation: $(1)^2+(2)^2-5=0$
Equation of circle $=1+5-5=0$
This represents that the point lies on the circumference of a circle, and hence only one tangent can be drawn from (1, 2).

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MCQ 1021 Mark

On the parabola $y=x^2$, the point least distant from the straight line $y=2 x-4$ is:

✓
(1, 1)
B
(1, 0)
C
(1, -1)
D
(0, 0)

Answer

Correct option: A.

(1, 1)

(1, 1)

Solution:
Given, parabola is $y=x^2 \ldots .$. (i)
and straight line is $\mathrm{y}=2 \mathrm{x}-4 \ldots$... (ii)
From equations (i) and (ii), we get
$x^2-2 x-4=0$
$\Rightarrow 2 x-2=0$
$\Rightarrow x=1$
From equation (i), we have $y=1$
The point least distant from the line is $(1,1)$.

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MCQ 1031 Mark

The equation of parabola whose focus is $(3,0)$ and directrix is $3 x+4 y=1$ is:

A
$16 x^2-9 y^2-24 x y-144 x+8 y+224=0$
B
$16 x^2+9 y^2-24 x y-144 x+8 y-224=0$
C
$16 x^2+9 y^2-24 x y-144 x-8 y+224=0$
✓
$16 x^2+9 y^2-24 x y-144 x+8 y+224=0$

Answer

Correct option: D.

$16 x^2+9 y^2-24 x y-144 x+8 y+224=0$

$16 x^2+9 y^2-24 x y-144 x+8 y+224=0$

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MCQ 1041 Mark

If the equation of a circle is $\lambda\text{x}^2+(2\lambda-3)\text{y}^2-4\text{x}+6\text{y}-1=0,$ then the coordinates of centre are:

A
$\Big(\frac{4}{3},\ -1\Big)$
✓
$\Big(\frac{2}{3},\ -1\Big)$
C
$\Big(\frac{-2}{3},\ 1\Big)$
D
$\Big(\frac{2}{3},\ 1\Big)$

Answer

Correct option: B.

$\Big(\frac{2}{3},\ -1\Big)$

$\Big(\frac{2}{3},\ -1\Big)$

Solution:
To find the centre:
Coefficient of $\text{x}^2$ = Coefficient of $\text{y}^2$
$\therefore\lambda=2\lambda-3\Rightarrow\lambda=3$
Therefore, the given equation can be rewritten as $3\text{x}^2+3\text{y}^2-4\text{x}+6\text{y}-1=0.$
$\therefore\text{x}^2+\text{y}^2-\frac{4}{3}\text{x}+2\text{y}-\frac{1}{3}=0$
Thus, the coordinates of the centre is $\Big(\frac{2}{3},\ -1\Big).$

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MCQ 1051 Mark

Choose the correct answer. The distance between the foci of a hyperbola is 16 and its eccentricity is 2. Its equation is:

✓
$\text{x}^2-\text{y}^2=32$
B
$\frac{\text{x}^2}{4}-\frac{\text{y}^2}{9}=1$
C
$2\text{x}-3\text{y}^2=7$
D
none of these.

Answer

Correct option: A.

$\text{x}^2-\text{y}^2=32$

We know that distance between the foci = 2ae
$\therefore\ 2\text{ae}=16$
$\Rightarrow\text{ae}=8$
Given that $\text{e}=\sqrt{2}$
$\therefore\ \sqrt{2}\text{a}=8$
$\Rightarrow\text{a}=4\sqrt{2}$
Now, $\text{b}^2=\text{a}^2(\text{e}^2-1)$
$\Rightarrow\text{b}^2=32(32-1)$
$\Rightarrow\text{b}^2=32$
So, the equation of the hyperbola is,
$\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}=1$
$\Rightarrow\frac{\text{x}^2}{32}-\frac{\text{y}^2}{32}=1$
$\Rightarrow\text{x}^2-\text{y}^2=32$

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MCQ 1061 Mark

The length of the latus-rectum of the parabola $y^2 + 8x − 2y + 17 = 0$ is

A
2
B
4
✓
8
D
16

Answer

Correct option: C.

Solution:
$y^2+8 x-2 y+17=0$
$\Rightarrow(y-1)^2-1+8 x+17=0$
$\Rightarrow(y-1)^2+8 x+16=0$
$\Rightarrow(y-1)^2=-8(x+2)$
Let $X = x +2, Y = y -1$
$\therefore Y^2=-8 X$
Comparing with $y ^2=4 ax$ :
$a=2$
Length of the latus rectum $=4 a=8$ units

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MCQ 1071 Mark

Centre of circle whose normals are $x^2-2 x y-3 x+6 y=0$, is:

✓
$\big(3, \frac{3}{2}\big)$
B
$\big(3, -\frac{3}{2}\big)$
C
$\big(\frac{3}{2},3\big)$
D
None of these

Answer

Correct option: A.

$\big(3, \frac{3}{2}\big)$

$\big(3, \frac{3}{2}\big)$

Solution:
$x^2-2 x y-3 x+6 y=0$
$\Rightarrow(x-3)(x-2 y)=0$
$\Rightarrow \mathrm{x}=3$ and $\mathrm{x}=2 \mathrm{y}$ are two normals.
The intersection point of these two normals will be the centre of the circle.
$\therefore \text { for } x=3$
$\Rightarrow y=\frac{x}{2}=\frac{3}{2}$
The intersection point is $\left(3, \frac{3}{2}\right)$ the centre of the given circle is $\left(3, \frac{3}{2}\right)$

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MCQ 1081 Mark

Equation of the directrix of the parabola $x^2=4 a y$ is:

A
x = -a
B
x = a
✓
y = -a
D
y = a

Answer

Correct option: C.

y = -a

y = -a

Solution:
Given, parabola $x^2=4 a y$
Now, its equation of directrix $=y=-a$

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MCQ 1091 Mark

Choose the correct answer. If e is the eccentricity of the ellipse $\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1(\text{a}<\text{b}),$ then:

A
$b^2=a^2\left(1-e^2\right)$
✓
$a^2=b^2\left(1-e^2\right)$
C
$a^2=b^2\left(e^2-1\right)$
D
$b^2=a^2\left(e^2-1\right)$

Answer

Correct option: B.

$a^2=b^2\left(1-e^2\right)$

$a^2=b^2\left(1-e^2\right)$

Solution:
Given equation is $\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1(\text{a}<\text{b})$
$\therefore\text{ Eccentricity e}=\sqrt{1-\frac{\text{a}^2}{\text{b}^2}}$
$\Rightarrow\text{e}^2=1-\frac{\text{a}^2}{\text{b}^2}$
$\Rightarrow\frac{\text{a}^2}{\text{b}^2}=(1-\text{e}^2)$
$\Rightarrow\text{a}^2=\text{b}^2(1-\text{e}^2)$

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MCQ 1101 Mark

The eccentricity of an ellipse is:

A
e = 1
B
e < 1
C
e > 1
✓
0 < e < 1

Answer

Correct option: D.

0 < e < 1

The eccentricity of an ellipse e $=(1-\frac{\text{a}^2}{\text{b}^2})$ and 0 < e < 1

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MCQ 1111 Mark

If the circles $x^2+y^2=a$ and $x^2+y^2-6 x-8 y+9=0$, touch externally, then $a=$

✓
1
B
-1
C
21
D
16

Answer

Correct option: A.

Solution:
$x^2+y^2=a$
And, $x^2+y^2-6 x-8 y+9=0$......(2)
Let circles (1) and (2) touch each other at point $P$.
The centre of the circle $x^2+y^2=a, 0$, is $(0,0)$.
The centre of the circle $x^2+y^2-6 x-8 y+9=0, C_1$, is $(3,4)$.
Also, radius of circle (1) $=\sqrt{\text{a}}=\text{OP}$
Radius of circle (2) $\sqrt{9+16-9}=4=\text{C}_1\text{P}$
From figure, we have:
$\Rightarrow\sqrt{3^2+4^2}=4+\sqrt{\text{a}}$
$\Rightarrow5=4+\sqrt{\text{a}}$
$\Rightarrow\text{a}=1$

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MCQ 1121 Mark

The length of the latus-rectum of the parabola $x^2-4 x-8 y+12=0$ is

A
4
B
6
✓
8
D
10

Answer

Correct option: C.

Solution:
Given:
$x^2-4 x-8 y+12=0$
$(x-2)^2-8 y+8=0$
$(x-2)^2=8 y-8=8(y-1)$
Let $X=x-2, Y=y-1$
$\therefore \mathrm{X}^2=8 \mathrm{Y}$
$\therefore$ Length of the latus rectum $=4 a=8$ units

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MCQ 1131 Mark

Equation of the ellipse in its standard form is $\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}=1$:

A
True
✓
False
C
Nither
D
Either

Answer

Correct option: B.

False

Equation of ellipse in standard form is
$\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}=1$

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MCQ 1141 Mark

If the parabola $y^2=4$ ax passes through the point $(3,2)$, then the length of its latusrectum is:

A
$\frac{2}{3}$
✓
$\frac{4}{3}$
C
$\frac{1}{3}$
D
$4$

Answer

Correct option: B.

$\frac{4}{3}$

$\frac{4}{3}$

Solution:
Since, the parabola $y^2=4$ ax passes through the point $(3,2)$
$\Rightarrow 2^2=4 a \times 3$
$\Rightarrow 4=12 a$
$\Rightarrow a=\frac{4}{12}$
$\Rightarrow a=\frac{1}{3}$
So, the length of latusrectum $=4 \mathrm{a}=4 \times\left(\frac{1}{3}\right)=\frac{4}{3}$

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MCQ 1151 Mark

The equation $16 x^2+y^2+8 x y-74 x-78 y+212=0$ represents

A
A circle
✓
A parabola
C
An ellipse
D
A hyperbola

Answer

Correct option: B.

A parabola

a parabola

Solution:
Comparing the given equation with $a x^2+b y^2+2 h x y+2 g x+2 f y+c=0$, we get:
$a=16, b=1, h=4$
We have: $h^2=16=a b$
Thus, the given equation represents a parabola.

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MCQ 1161 Mark

If a be the radius of a circle which touches x-axis at the origin, then its equation is:

A
$\text{x}^2 + \text{y}^2 + \text{ax} = 0$
B
$\text{x}^{2} + \text{y}^{2} \underline{+} 2\text{ya} = 0$
✓
$\text{x}^{2} + \text{y}^{2} \underline{+} 2\text{xa} = 0$
D
$\text{x}^2 + \text{y}^2 + \text{ya} = 0$

Answer

Correct option: C.

$\text{x}^{2} + \text{y}^{2} \underline{+} 2\text{xa} = 0$

$\text{x}^{2} + \text{y}^{2} \underline{+} 2\text{xa} = 0$

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MCQ 1171 Mark

Find the Center of circle $x^2+y^2-4 x-8 x+25=0:$

✓
(2, 4)
B
(-2, -4)
C
(4, 2)
D
(-4, -2)

Answer

Correct option: A.

(2, 4)

(2, 4)

Solution:
The general equation of center of circle $x^2+y^2+2 g x+2 f y+c=0$ is ( $-g,-f$ )
So, the center of circle $x^2+y^2-4 x,-8 x+25=0$ is $(2,4)$

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MCQ 1181 Mark

The equation of the circle drawn with the two foci of $\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1$ end-point of a diameter is

A
$\text{x}^2+\text{y}^2=\text{a}^2+\text{b}^2$
B
$\text{x}^2+\text{y}^2=\text{a}^2$
C
$\text{x}^2+\text{y}^2=2\text{a}^2$
✓
$\text{x}^2+\text{y}^2=\text{a}^2-\text{b}^2$

Answer

Correct option: D.

$\text{x}^2+\text{y}^2=\text{a}^2-\text{b}^2$

We have $\text{r}=\text{ae}$
Let the equation of the circle be $\text{x}^2+\text{y}^2=\text{r}^2.$
Now, $\text{x}^2+\text{y}^2=\text{a}^2\text{e}^2$ $(\because\text{r}=\text{ae})$
$\Rightarrow\text{x}^2+\text{y}^2=\text{a}^2\Big(1-\frac{\text{b}^2}{\text{a}^2}\Big)$ $\bigg(\because\text{e}=\sqrt{1-\frac{\text{b}^2}{\text{a}^2}}\bigg)$
$\Rightarrow\text{x}^2+\text{y}^2=\text{a}^2-\text{b}^2$
$\therefore\ $The required equation of the circle $\text{x}^2+\text{y}^2=\text{a}^2-\text{b}^2.$

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MCQ 1191 Mark

The foci of the hyperbola $9x^2 − 16y^2 = 144$ are

A
$(\pm4,0)$
B
$(0,\pm4)$
✓
$(\pm5,0)$
D
$(0,\pm5)$

Answer

Correct option: C.

$(\pm5,0)$

$(\pm5,0)$

Solution:
The equation of the hyperbola is given below:
$9x^2 − 16y^2 = 144$
This equation can be rewritten in the following way:
$\frac{9\text{x}^2}{144}-\frac{16\text{y}^2}{144}=1$
$\Rightarrow\frac{\text{x}^2}{16}-\frac{\text{y}^2}{9}=1$
This is the standard equation of a hyperbola, where $a^2 = 16$ and $b^2 = 9$.
The eccentricity is calculated in the following way:
$b^2 = a^2(e^2 − 1)$
$\Rightarrow 9 = 16(e^2 − 1)$
$\Rightarrow\frac{9}{16}=\text{e}^2-1$
$\Rightarrow\text{e}=\frac{5}{4}$
$\text{Foci}=(\pm\text{ae},0)=(\pm5,0)$

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MCQ 1201 Mark

The eccentricity of the ellipse, if the minor axis is equal to the distance between the foci, is:

A
$\frac{\sqrt{3}}{2}$
B
$\frac{2}{\sqrt{3}}$
✓
$\frac{1}{\sqrt{2}}$
D
$\frac{\sqrt{2}}{3}$

Answer

Correct option: C.

$\frac{1}{\sqrt{2}}$

According to the question, the minor axis is equal to the distance between the foci.
i.e. $2\text{b}=2\text{ae}$
$\text{e}=\frac{\text{b}}{\text{a}}\ \dots(1)$
Now, $\text{e}=\sqrt{1-\frac{\text{b}^2}{\text{a}^2}}$
$\Rightarrow\text{e}=\sqrt{1-\text{e}^2}$ $\Big[\text{From}\ (1)\Big]$
On squaring both sides, we get:
$\text{e}^2=1-\text{e}^2$
$\Rightarrow2\text{e}^2=1$
$\Rightarrow\text{e}^2=\frac{1}{2}$
$\Rightarrow\text{e}=\frac{1}{\sqrt{2}}$

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MCQ 1211 Mark

Find the center-radius form of the equation of the circle with center $(4,0)$ and radius 7 :

✓
$(x-4)^2+y^2=49$
B
$x^2+(y+4)^2=7$
C
$x^2+(y-4)^2=7$
D
$(x+4)^2+y^2=49$

Answer

Correct option: A.

$(x-4)^2+y^2=49$

$(x-4)^2+y^2=49$

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MCQ 1221 Mark

If the length of the tangent from the origin to the circle centered at $(2,3)$ is 2 then the equation of the circle is:

A
$(x+2)^2+(y-3)^2=3^2$
B
$(x-2)^2+(y+3)^2=3^2$
✓
$(x-2)^2+(y-3)^2=3^2$
D
$(x+2)^2+(y+3)^2=3^2$

Answer

Correct option: C.

$(x-2)^2+(y-3)^2=3^2$

$(x-2)^2+(y-3)^2=3^2$

Solution:
Radius of the circle $= \sqrt{{(2 - 0)^2 + (3 - 0)^2 - 2^2}}$
$=\sqrt{(4 + 9 - 4)}$
$= \sqrt{9}$
= 3
So, the equation of the circle = $(x-2)^2+(y-3)^2=3^2$

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MCQ 1231 Mark

The radius of the circle passing through the point (6, 2) and two of whose diameters are x + y = 6 and x + 2y = 4 is:

A
4
B
6
C
20
✓
$\sqrt { 20 }$

Answer

Correct option: D.

$\sqrt { 20 }$

Point of intersection of the given diameters is (8, -2) which is the centre of the circle.
Also the circle pass through the point (6, 2) so the radius is.
$=\sqrt{ (8-6)^2+(-2-2)^2}=\sqrt{20}$

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MCQ 1241 Mark

If the vertices of a triangle are (2, -2), (-1, -1) and (5, 2) then the equation of its circumcircle is:

A
$x^2+y^2+3 x+3 y+8=0$
✓
$x^2+y^2-3 x-3 y-8=0$
C
$x^2+y^2-3 x+3 y+8=0$
D
None of these

Answer

Correct option: B.

$x^2+y^2-3 x-3 y-8=0$

$x^2+y^2-3 x-3 y-8=0$

Solution:
To find circumcentre we write the equation of perpendicular bisectors of two sides and find their intersection,
3x - y - 3 = 0 and 6x + 8y - 21 = 0
Their intersection point is $\big(\frac{3}{2},\frac{3}{2}\big)$
Radius of circumcircle = Distance of $\big(\frac{3}{2},\frac{3}{2}\big)$
from (2,-2) or any other vertex $=\frac{5}{\sqrt2}$
So equation of circle $=(\text{x}-\frac{3}{2})^2+(\text{y}-\frac{3}{2})^2 = \frac{25}{2}$

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MCQ 1251 Mark

The focus of parabola $y^2=8 x$ is:

✓
(2, 0)
B
(-2, 0)
C
(0, 2)
D
(0, -2)

Answer

Correct option: A.

(2, 0)

(2, 0)

Solution:
Given, $\mathrm{y}^2=8 \mathrm{x}$
General equation is $\mathrm{y}^2=4 a x$
Now, $4 \mathrm{a}=8$
$\Rightarrow a=2$
Now, focus $=(a, 0)=(2,0)$

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MCQ 1261 Mark

The equation of the circle passing through the origin which cuts off intercept of length 6 and 8 from the axes is:

A
$x^2+y^2-12 x-16 y=0$
B
$x^2+y^2+12 x+16 y=0$
C
$x^2+y^2+6 x+8 y=0$
✓
$x^2+y^2-6 x-8 y=0$

Answer

Correct option: D.

$x^2+y^2-6 x-8 y=0$

$x^2+y^2-6 x-8 y=0$

Solution:
The centre of the required circle is $\left(\frac{6}{2}, \frac{8}{2}\right)=(3,4)$.
The radius of the required circle is $\sqrt{3^2+4^2}=\sqrt{25}=5$
Hence, the equation of the circle is as follows:
$(x-3)^2+(y-4)^2=52$
$\Rightarrow x^2+y^2-6 x-8 y=0$

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MCQ 1271 Mark

The vertex of the parabola $y^2-4 y-x+3=0$ is:

A
$(-1,3)$
✓
$(-1,2)$
C
$(2,-1)$
D
$(3,-1)$

Answer

Correct option: B.

$(-1,2)$

$(-1,2)$

Solution:
$y^2-4 y-x+3=0$
$(y-2)^2-4-x+3=0$
$(y-2)^2=(x+1)$
$\therefore$ Vertex of the parabola $=(-1,2)$

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MCQ 1281 Mark

Which of the following equations represents a parabola:

A
$(x - y)^3 = 3$
B
$\frac{\text{x}}{\text{y}}-\frac{\text{y}}{\text{x}}=0$
✓
$\frac{\text{x}}{\text{y}}+\frac{\text{4}}{\text{x}}=0$
D
$(x + y)^2+ 3 = 0$

Answer

Correct option: C.

$\frac{\text{x}}{\text{y}}+\frac{\text{4}}{\text{x}}=0$

$\frac{\text{x}}{\text{y}}+\frac{\text{4}}{\text{x}}=0$

Solution:
We know that the general equation of parabola is
$y^2=4 a x$
$y^2=-4 a x$
$x^2=4 a x$
$x^2=-4 a x$
From option (a),
$(x-y)^3=3$
It is not represent the parabola.
From option (b),
$\frac{x}{y}-\frac{y}{x}=0$
$x^2=y^2$
It is not represent the parabola.
From option (c),
$\frac{x}{y}+\frac{4}{x}=0$
$x^2+4 y=0$
$x^2=-4 y$
So, this is represented the parabola.
From option (d),
$(x+y)^3+3=0$
It is not represent the parabola.

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MCQ 1291 Mark

Choose the correct answer. Equation of the hyperbola with eccentricty $\frac{3}{2}$ and foci at $(\pm2,0)$ is:

✓
$\frac{\text{x}^2}{4}-\frac{\text{y}^2}{5}=\frac{4}{9}$
B
$\frac{\text{x}^2}{9}-\frac{\text{y}^2}{9}=\frac{4}{9}$
C
$\frac{\text{x}^2}{4}-\frac{\text{y}^2}{9}=1$
D
none of these.

Answer

Correct option: A.

$\frac{\text{x}^2}{4}-\frac{\text{y}^2}{5}=\frac{4}{9}$

Given that $\text{e}=\frac{3}{2}$
and foci $=(\pm\text{ae},0)=(\pm2,0)$
$\therefore\ \text{ae}=2$
$\text{a}\times\frac{3}{2}=2$
$\Rightarrow\text{a}=\frac{4}{3}$
Now we know that $\text{b}^2=\text{a}^2(\text{e}^2-1)$
$\text{b}^2=\frac{16}{9}\Big(\frac{9}{4}-1\Big)$
$\Rightarrow\text{b}^2=\frac{16}{9}\times\frac{5}{4}$
$\Rightarrow\text{b}^2=\frac{20}{9}$
So, the equation of the hyperbola is,
$\frac{\text{x}^2}{\big(\frac{4}{3}\big)^2}-\frac{\text{y}^2}{\frac{20}{9}}=1$
$\Rightarrow\frac{9\text{x}^2}{16}-\frac{9\text{y}^2}{20}=1$
$\Rightarrow\frac{\text{x}^2}{16}-\frac{\text{y}^2}{20}=\frac{1}{9}$
$\Rightarrow\frac{\text{x}^2}{4}-\frac{\text{y}^2}{5}=\frac{4}{9}$

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MCQ 1301 Mark

Coordinates of centre and radius of the circle $(x-3)^2+(y+4)^2=25$ are respectively:

A
$(3,4), 25$
B
$(-3,4), 5$
✓
$(3,-4), 5$
D
$(3,-4), 25$

Answer

Correct option: C.

$(3,-4), 5$

$(3,-4), 5$

Solution:
$(x-3)^2+(y+4)^2=25$
$(x-3)^2+(y-(-4))^2-(5)^2$
$(x-4)^2+(y-k)^2-(r)^2$
$r=5(h, k)=(3,-4)$

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MCQ 1311 Mark

The straight line $y=m x+c$ cuts the circle $x^2+y^2=a^2$ in real points if:

A
$\sqrt{{\text{a}^2 × (1 + \text{m}^2)} < \text{c}}$
B
$\sqrt{{\text{a}^2 × (1 - \text{m}^2)} < \text{c}}$
✓
$\sqrt{{\text{a}^2 × (1 + \text{m}^2)} > \text{c}}$
D
$\sqrt{{\text{a}^2 × (1 - \text{m}^2)} > \text{c}}$

Answer

Correct option: C.

$\sqrt{{\text{a}^2 × (1 + \text{m}^2)} > \text{c}}$

$\sqrt{{\text{a}^2 × (1 + \text{m}^2)} > \text{c}}$

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MCQ 1321 Mark

The diameter of a circle described by $\text{9x}^{2}+\text{9y}^2=16$ is:

A
$\frac{16}{9}$
B
$\frac{4}{3}$
C
$4$
✓
$\frac{8}{3}$

Answer

Correct option: D.

$\frac{8}{3}$

$\frac{8}{3}$

Solution:
Equation of circle is $\text{9x}^{2}+\text{9y}^2=16$
$\Rightarrow\text{x}^2+\text{y}^2=\frac{16}{9}$
$\Rightarrow\text{x}^2+\text{y}^2=(\frac{4}{3})^2$
$\therefore$ Radius of the circle $=\frac{4}{3}$
$\therefore$ Diameter of the circle $=\frac{4\times2}{3} = \frac{8}{3}$

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MCQ 1331 Mark

The directrix of the parabola $x^2-4 x-8 y+12=0$ is

A
y = 0
B
x = 1
✓
y = -1
D
x = -1

Answer

Correct option: C.

y = -1

y = -1

Solution:
Given:
$x^2-4 x-8 y+12=0$
$\Rightarrow(x-2)^2-4-8 y+12=0$
$\Rightarrow(x-2)^2=8 y-8$
$\Rightarrow(x-2)^2=8(y-1) \backslash$
Putting $X=x-2, Y=y-1$ :
$X^2=8 Y$
Comparing with $\mathrm{X}^2=4 \mathrm{aY}$ :
$a=2$
Equation of the directrix:
$Y=-a$
$\Rightarrow Y=-2$
$\Rightarrow y-1=-2$
$\Rightarrow y=-2+1$
$\Rightarrow y=-1$

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MCQ 1341 Mark

The locus of a planet orbiting around the sun is:

A
A circle
B
A straight line
C
A semicircle
✓
An ellipse

Answer

Correct option: D.

An ellipse

It is a fact & proof of it can be seen from higher education physics books.

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MCQ 1351 Mark

If the length of the major axis of an ellipse is three times the length of the minor axis then its eccentricity is:

A
$\frac{1}{3}$
B
$\frac{1}{\sqrt{3}}$
C
$\frac{1}{\sqrt{2}}$
✓
$\frac{2\sqrt{2}}{\sqrt{2}}$

Answer

Correct option: D.

$\frac{2\sqrt{2}}{\sqrt{2}}$

$\frac{2\sqrt{2}}{\sqrt{2}}$

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MCQ 1361 Mark

The equation of the parabola whose focus is (1, -1) and the directrix is x + y + 7 = 0 is

A
$x^2+y^2-2 x y-18 x-10 y=0$
B
$x^2-18 x-10 y-45=0$
C
$x^2+y^2-18 x-10 y-45=0$
✓
$x^2+y^2-2 x y-18 x-10 y-45=0$

Answer

Correct option: D.

$x^2+y^2-2 x y-18 x-10 y-45=0$

$x^2+y^2-2 x y-18 x-10 y-45=0$

Solution:
Let P (x, y) be any point on the parabola whose focus is S (1, -1) and the directrix is x + y+ 7 = 0.

Draw PM perpendicular to x + y + 7 = 0.
Then, we have:
SP = PM
$\Rightarrow SP^2 = PM^2$
$\Rightarrow\ (\text{x} - 1)^2+ (\text{y} + 1)^2= \Big(\frac{\text{x+y+7}}{\sqrt{1+1}}\Big)^2$
$\Rightarrow\ (\text{x} - 1)^2+ (\text{y} + 1)^2= \Big(\frac{\text{x+y+7}}{\sqrt{2}}\Big)^2$
$\Rightarrow\ 2(\text{x}^2+1-2\text{x}+\text{y}^2+1+2\text{y})\\ \ \ =\text{x}^2+\text{y}^2+49+2\text{xy}+14\text{y}+14\text{x}$
$\Rightarrow\ (2\text{x}^2+2-4\text{x}+2\text{y}^2+2+4\text{y})\\ \ \ =\text{x}^2+\text{y}^2+49+2\text{xy}+14\text{y}+14\text{x}$
$\Rightarrow\ \text{x}^2+\text{y}^2-45-10\text{y}-2\text{xy}-18\text{x}=0$
Hence, the required equation is $x^2 + y^2 - 2xy - 18x - 10y - 45 = 0$.

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MCQ 1371 Mark

If the lines $3 x-4 y-7=0$ and $2 s-3 y-5=0$ are two diameters of a circle of area $49 \pi$ square units, the equation of the circle is:

A
$x^2+y^2+2 x-2 y-62=0$
B
$x^2+y^2-2 x+2 y-62=0$
✓
$x^2+y^2-2 x+2 y-47=0$
D
$x^2+y^2+2 x-2 y-47=0$

Answer

Correct option: C.

$x^2+y^2-2 x+2 y-47=0$

$x^2+y^2-2 x+2 y-47=0$

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MCQ 1381 Mark

One of the diameters of the circle $x^2+y^2-12 x+4 y+6=0$ is given by:

A
x + y = 0
✓
x + 3y = 0
C
x = y
D
3x + 2y = 0

Answer

Correct option: B.

x + 3y = 0

x + 3y = 0

Solution:
The coordinate of the centre of the circle $x^2+y^2-12 x+4 y+6=0$ are $(6,-2)$
Clearly, the line $x+3 y$ passes through this point.
$x+3 y=0$ is a diameter of the given circle.

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MCQ 1391 Mark

The latus-rectum of the conic $3\text{x}^2+4\text{y}^2-6\text{x}+8\text{y}-5=0$ is:

✓
$3$
B
$\frac{\sqrt{3}}{2}$
C
$\frac{2}{\sqrt{3}}$
D
$\text{none of these}$

Answer

Correct option: A.

$3$

$3\text{x}^2+4\text{y}^2-6\text{x}+8\text{y}-5=0$
$\Rightarrow3(\text{x}^2-2\text{x})+4(\text{y}^2+2\text{y})=5$
$\Rightarrow3(\text{x}^2-2\text{x}+1)+4(\text{y}^2+2\text{y}+1)=5+3+4$
$\Rightarrow3(\text{x}-1)^2+4(\text{y}+1)^2=12$
$\frac{(\text{x-1})^2}{4}+\frac{(\text{y}+1)^2}{3}=1$
So, $\text{a}=2$ and $\text{b}=\sqrt{3}$
$\therefore\ $Latus rectum $=\frac{2\text{b}^2}{\text{a}}$
$\\=2\frac{\big[\sqrt{3}\big]^2}{2}\\=3$

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MCQ 1401 Mark

Choose the correct answer. Equation of the circle with centre on the y-axis and passing through the origin and the point $(2, 3)$ is:

✓
$x^2+y^2+13 y=0$
B
$3 x^2+3 y^2+13 x+3=0$
C
$6 x^2+6 y^2-13 x=0$
D
$x^2+y^2+13 x+3=0$

Answer

Correct option: A.

$x^2+y^2+13 y=0$

$x^2+y^2+13 y=0$

Solution:
Let the equation of the circle be,
$(x - h)^2 + (y - k)^2 = r^2$
Let the centre be (0, a)
$\therefore$ Radius r = a
So, the equation of the circle is
$(x - 0)^2 + (y - a)^2 = a^2$
$\Rightarrow x^2 + (y - a)^2 = a^2$
$\Rightarrow x^2 + y^2 + a^2 - 2ay = a^2$
$\Rightarrow x^2 + y^2 - 2ay = 0 .....(i)$
(image)
Now, CP = r
$\Rightarrow\sqrt{(2-0)^2+(3-\text{a}^2)}=\text{a}$
$\Rightarrow\sqrt{4+9+\text{a}^2-6\text{a}}=\text{a}$
$\Rightarrow\sqrt{13+\text{a}^2-6\text{a}}=\text{a}$
$\Rightarrow13+\text{a}^2-6\text{a}=\text{a}^2$
$\Rightarrow13-6\text{a}=0$
$\therefore\ \text{a}=\frac{13}{6}$
Putting the value of a in eq. (i) we get
$\text{x}^2+\text{y}^2-2\Big(\frac{13}{6}\Big)\text{y}=0$
$\Rightarrow3\text{x}^2+3\text{y}^2-3\text{y}=0$
Note: (a) option is correct and is should be (dout solution)

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MCQ 1411 Mark

If (x, 3) and (3, 5) are the extremities of a diameter of a circle with centre at (2, y), then the values of x and y are:

A
(3, 1)
B
x = 4, y = 1
C
x = 8, y = 2
✓
None of these

Answer

Correct option: D.

None of these

The end points of the diameter of a circle are (x, 3) and (3, 5).
According to the question, we have:
$\frac{\text{x}+3}{2}=2,\ \text{y}=\frac{5+3}{2}$
$\Rightarrow\text{x}=1,\ \text{y}=4$

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MCQ 1421 Mark

If the equation $(4 a-3) x^2+a y^2+6 x-2 y+2=0$ represents a circle, then its centre is:

A
$(3,-1)$
B
$(3,1)$
✓
$(-3,1)$
D
None of these

Answer

Correct option: C.

$(-3,1)$

$(-3, 1)$

Solution:
If the equation $(4 a-3) x^2+a y^2+6 x-2 y+2=0$ represents a circle, then we have:
Coefficient of $\mathrm{x}^2=$ Coefficient of $\mathrm{y}^2$
$\Rightarrow 4 a-3=a$
$\Rightarrow a=1$
$\therefore$ Equation of the circle $=x^2+y^2+6 x-2 y+2=0$
Thus, the coordinates of the centre is $(-3,1)$.

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MCQ 1431 Mark

The length of latus rectum of the parabola $(x-2 a)^2+y^2=x^2$ is:

A
2a
B
3a
C
6a
✓
4a

Answer

Correct option: D.

Solution:
We have, $(x-2 a)^2+y^2=x^2$
$x^2-4 a x+4 a^2+y^2=x^2$
$y^2=4 a x-4 a^2=4 a(x-a)$
Comparing it with standard parabola $\mathrm{Y}^2=4 \mathrm{bX}$
$Y=y, X=x-a, b=a$
We know length of latus rectum of parabola $Y^2=4 b X$ is $4 b$ length of latus rectum of given parabola is $=4 \times a=4 a$

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MCQ 1441 Mark

The equation of the conic $9 x^2-16 y^2=144$ is

✓
$\frac{5}{4}$
B
$\frac{4}{3} $
C
$\frac{4}{5}$
D
$\sqrt7$

Answer

Correct option: A.

$\frac{5}{4}$

$\frac{5}{4}$

Solution:
Standard form of a hyperbola $=\frac{\mathrm{x}^2}{16}-\frac{\mathrm{y}^2}{9}=1$
Here, $a^2=16$ and $y^2=9$
The eccentricity is calculated in the following way:
$b^2=a^2\left(e^2-1\right)$
$\Rightarrow 9=16\left(e^2-1\right)$
$\Rightarrow e^2-1=\frac{9}{16}$
$\Rightarrow e^2=\frac{25}{16}$
$\Rightarrow e=\frac{5}{4}$

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MCQ 1451 Mark

An ambulance company provides services within an 80 mile radius of their headquarters If this service area is represented graphically with the headquarters located at the coordinates (0, 0) what is the equation that represents the service area:

A
$x^2+y^2=80$
B
$(x-0)^2+(y-0)^2=80$
C
$x^2+y^2=1600$
✓
$x^2+y^2=6400$

Answer

Correct option: D.

$x^2+y^2=6400$

$x^2+y^2=6400$

Solution:
The general equation of a circle with center at $(a, b)$ and radius $r$ is $(x-a)^2+(y-b)^2=r^2$ so substituting the values we get the circle equation as $x^2+y^2=80^2=6400$

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MCQ 1461 Mark

The equation of the parabola with focus (0, 0) and directrix x + y = 4 is

✓
$x^2+y^2-2 x y+8 x+8 y-16=0$
B
$x^2+y^2-2 x y+8 x+8 y=0$
C
$x^2+y^2+8 x+8 y-16=0$
D
$x^2-y^2+8 x+8 y-16=0$

Answer

Correct option: A.

$x^2+y^2-2 x y+8 x+8 y-16=0$

$x^2+y^2-2 x y+8 x+8 y-16=0$

Solution:
Let P (x, y) be any point on the parabola whose focus is S (0, 0) and the directrix is x + y = 4.

Draw PM perpendicular to x + y = 4.
Then, we have:
SP = PM
$\Rightarrow SP^2 = PM^2$
$\Rightarrow\ (\text{x}-0)^2+(\text{y}-0)^2=\Big(\frac{\text{x+y}-4}{\sqrt2}\Big)^2$
$\Rightarrow\ \text{x}^2+\text{y}^2=\Big(\frac{\text{x+y}-4}{\sqrt2}\Big)^2$
$\Rightarrow 2x^2 + 2y^2 = x^2 + y^2 + 16 + 2xy - 8x - 8y$
$\Rightarrow x^2 + y^2 - 2xy + 8x + 8y - 16 = 0$

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MCQ 1471 Mark

The length of the latus-rectum of the parabola $4 y^2+2 x-20 y+17=0$ is

✓
$3$
B
$6$
C
$\frac{1}{2}$
D
$9$

Answer

Correct option: A.

$3$

$\frac{1}{2}$

Solution:
Given:
$4 y^2+2 x-20 y+17=0$
$\Rightarrow\ \text{y}^2+\frac{\text{x}}{2}-5\text{y}+\frac{17}{4}=0$
$\Rightarrow\ \Big(\text{y}-\frac{5}{2}\Big)^2+\frac{\text{x}}{2}-2=0$
$\Rightarrow\ \Big(\text{y}-\frac{5}{2}\Big)^2=-1\Big(\frac{\text{x}}{2}-2\Big)$
$\Rightarrow\ \Big(\text{y}-\frac{5}{2}\Big)^2=\frac{-1}{2}(\text{x}-4)$
$\text{Let }\text{X}=\text{x}-4,\ \text{Y}=\text{y}-\frac{5}{2}$
$\therefore\ \text{Y}^2=\frac{-\text{X}}{2}$
$\therefore$ Length of the latus rectum $=\ 4\text{a}=\frac{1}{2}$ units

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MCQ 1481 Mark

The radius of the circle with center (0, 0) and which passes through (-6, 8) is:

A
5
✓
10
C
6
D
8

Answer

Correct option: B.

$\text{r}=\sqrt{(6)^2+(-8)=10}$

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MCQ 1491 Mark

Find the area of $x^2+y^2=49$:

✓
154
B
49
C
88
D
None

Answer

Correct option: A.

154

Solution:
Th eequation $x^2+y^2=49$ describes a circle with 7 as radius So the area is given as $\pi\text{r}^2$
$=\frac{22}{7}\times7^2=154$

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MCQ 1501 Mark

Choose the correct answer. The length of the latus rectum of the ellipse $3\text{x}^2+\text{y}^2=12$ is:

A
4
B
3
C
8
✓
$\frac{4}{\sqrt{3}}$

Answer

Correct option: D.

$\frac{4}{\sqrt{3}}$

$\frac{4}{\sqrt{3}}$

Solution:
$3\text{x}^2+\text{y}^2=12$
$\Rightarrow\frac{\text{x}^2}{4}+\frac{\text{y}^2}{12}=1$
$\therefore\text{ a}^2=4$
$\Rightarrow\text{a}=2$
and $\text{b}^2=12$
$\Rightarrow\text{b}=2\sqrt{3}$
Since b > a, length of latus rectum $=\frac{2\text{a}^2}{\text{b}}=\frac{2\times4}{2\sqrt{3}}=\frac{4}{\sqrt{3}}$

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MCQ 1511 Mark

The equation of the ellipse with focus $(-1, 1),$ directrix $x - y + 3 = 0$ and eccentricity $\frac{1}{2}$ is:

A
$7\text{x}^2+2\text{xy}+7\text{y}^2+10\text{x}+10\text{y}+7=0$
✓
$7\text{x}^2+2\text{xy}+7\text{y}^2+10\text{x}-10\text{y}+7=0$
C
$7\text{x}^2+2\text{xy}+7\text{y}^2+10\text{x}-10\text{y}-7=0$
D
$\text{None of these}$

Answer

Correct option: B.

$7\text{x}^2+2\text{xy}+7\text{y}^2+10\text{x}-10\text{y}+7=0$

Let $P(x,y)$ be any point on the ellipse whose focus and eccentricity are $S(-1,1)$ and $\text{e}=\frac{1}{2},$
respectively.Let $PM$ be the perpendicular from $P$ on the directrix.
Then $\text{SP}=\text{e}\times\text{PM}$
$\Rightarrow\text{SP}=\frac{1}{2}\times\text{PM}$
$\Rightarrow2\text{SP}=\text{PM}$
$\Rightarrow4(\text{SP})^2=\text{PM}^2$
$\Rightarrow4\Big[(\text{x}+1)^2+(\text{y}-1)^2\Big]=\bigg(\frac{\text{x}-\text{y}+3}{\sqrt{1^2+}(-1)^2}\bigg)^2$
$\Rightarrow4\big[\text{x}^2+1+2\text{x}+\text{y}^2+1-2\text{y}\big]\\=\frac{{\text{x}^2+\text{y}^2+9-2\text{xy}-6\text{y}+6\text{x}}}{2}$
$\Rightarrow8\text{x}^2+8+16\text{x}+8\text{y}^2+8-16\text{y}\\=\text{x}^2+\text{y}62+9-2\text{xy}-6\text{y}+6\text{x}$
$\therefore7\text{x}^2+7\text{y}^2+2\text{xy}-10\text{y}+10\text{x}+7=0$
This is the required equation of the ellipse.

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MCQ 1521 Mark

The difference of the focal distances of any point on the hyperbola is equal to

A
length of the conjugate axis.
B
eccentricity.
✓
length of the transverse axis.
D
Latus-rectum.

Answer

Correct option: C.

length of the transverse axis.

Let P(x,y) be any point on the hyperbola, and S, S' be the focus with coordinates $(\pm\text{ae},0).$
⇒ S'P − SP = 2a
Thus, the difference of the focal distances of any point on the hyperbola is equal to the length of the transverse axis.

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MCQ 1531 Mark

A point moves in a plane so that its distances PA and PB from two fixed points A and B in the plane satisfy the relation PA − PB = k (k ≠ 0), then the locus of P is

✓
A hyperbola.
B
A branch of the hyperbola.
C
A parabola.
D
An ellipse.

Answer

Correct option: A.

A hyperbola.

Let P(x, y) be any point on the hyperbola $\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2}=1.$
By definition, we have:
$\text{PA}=\text{e}\big(\text{x}-\frac{\text{a}}{\text{e}}\big)=\text{ex}-\text{a}$
$\text{and }\text{PB}=\text{e}\big(\text{x}+\frac{\text{a}}{\text{e}}\big)=\text{ex}+\text{a}$
$\therefore\text{PB}−\text{PA}=\text{(ex+a)}−\text{(ex}−\text{a})=\text{2a = k}$

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