M.C.Q (1 Marks)

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.

$\frac{3}{4}$

$\frac{\lambda(\text{x}+1)^2}{3}+\frac{(\text{y}+2)^2}{4}=1$
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 $

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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.

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

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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 $

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 $

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.

$(-1, 2)$

Let the coordinates of $P$ and $Q$ be $(at{_1}^2, 2at_1)$ and $(at{_2}^2, 2at_2)$, respectively.
Slope of $PQ =\frac{2\text{at}_2-2\text{at}_1}{\text{at}_2^2-\text{at}_1^2}...(1)$
But, the slope of $PQ$ is equal to the slope of $2x - y + 4 = 0.$
$\therefore$ Slope of $PQ=\frac{-2}{-1}=2$
From $(1),$
$\frac{2\text{at}_2-2\text{at}_1}{\text{at}_2^2-\text{at}_1^2}=2\ ...(2)$
Putting $4a = 8,$
$a = 2$
$\therefore$ Focus of the given parabola $= (a, 0) = (2, 0)$
Using equation $(2):$
$\frac{4(\text{t}_2-\text{t}_1)}{2(\text{t}_2^2-\text{t}_1^2)}=2$
$\frac{(\text{t}_2-\text{t}_2)}{(\text{t}_2^2-\text{t}_1^2)}=1$
$\Rightarrow\ \text{t}_1+\text{t}_2=1$
As, points $P$ and $Q$ lie on $2x - y + 4 = 0$
$\Rightarrow P(at{_1}^2, 2at_1)$ or $P(2t{_1}^2, 4t_1)$ lie on line $2x - y + 4 = 0$
$\Rightarrow 2(2t{_1}^2) - (4t_1) + 4 = 0$
$\Rightarrow t{_1}^2 - t_1 + 1 = 0 ...(3)$
Also, $Q(at{_2}^2, 2at_2)$ or $P(2t{_2}^2, 4t_2)$ lie on line $2x - y + 4 = 0$
$\Rightarrow 2(2t{_2}^2) - (4t_2) + 4 = 0$
$\Rightarrow t{_2}^2 - t_2 + 1 = 0 ...(4)$
Adding $(3)$ and $(4),$ we get,
$\Rightarrow t{_1}^2 - t_1 + 1 + t{_2}^2- t_2 + 1 = 0$
$\Rightarrow (t{_1}^2 + t{_2}^2) - (t_1 + t_2) + 2 = 0$
$\Rightarrow (t{_1}^2 +t{_2}^2) - 1 + 2 = 0 [t_1 + t_2 = 1$, proved above$]$
$\Rightarrow (t{_1}^2 + t{_2}^2) = -1$
Let $(x_1, y_1)$ be the mid$-$point of $PQ.$
Then, we have:
$\text{y}_1=\frac{2\text{at}_2+2\text{at}_1}{2}=2(\text{t}_1+\text{t}_2)=2$
And, $\text{x}_1=\frac{\text{at}_1^2+\text{at}_2^2}{2}=\text{t}_1^2+\text{t}_2^2=-1$
$\Rightarrow\ (\text{x}_1, \text{y}_1)=(-1, 2)$

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MCQ 81 Mark

The circle $x^2+ y^2- 3x - 4y + 2 = 0$ 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.

$(1, 0), (2, 0)$

$x^2+ y^2- 3x - 4y + 2 = 0$
$x-$axis will be cut when $y = 0$
put $y=0$
$x^2 - 3x + 2 = 0$
$(x - 2) (x - 1) = 0$
$x = 1, 2$
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.

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

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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.

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

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.

$\frac{4}{5}$

$\Rightarrow\frac{\text{x}^2}{25}+\frac{\text{y}^2}{9}=1$
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 $x + y = 3a,$ 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 $

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:
$\text{SP = PM}$
$\Rightarrow \ce{SP^2 = PM^2}$
$\Rightarrow\ (\text{x}-0)^2+(\text{y+a})^2=\Big(\frac{\text{x+y}-3\text{a}}{\sqrt2}\Big)^2$
$\Rightarrow\ \text{x}^2+(\text{y+a})^2=\Big(\frac{\text{x+y}-3\text{a}}{\sqrt2}\Big)^2$
$\Rightarrow\ 2\text{x}^2+2\text{y}^2+2\text{a}^2+4\text{ay}=\text{x}^2+\text{y}^2+9\text{a}^2+2\text{xy}-6\text{ax}-6\text{ay}$
$\Rightarrow\ \text{x}^2+\text{y}^2-7\text{a}^2+10\text{ay}+6\text{ax}=0$

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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, $\sqrt{(\text{x}-2)^{2}+\text{y}^{2}}+\sqrt{(\text{x}+2)^{2}+\text{y}^{2}}=5>\text{AB}$
$\Rightarrow PA + PB =$ 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.

$6$

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 $x^2+ y^2- 2x = 0$ is $AB.$ Equation of the circle with $AB$ 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 $

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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 $

Compare the equation with the standard form with center at $(h, k)$ 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
$y^2 = 2x - 4$
B
$x^2 = 4y - 8$
✓
$y^2 = 4x - 8$
D
None

Answer

Correct option: C.

$y^2 = 4x - 8$

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

Let the required equation of the circle be $(x - h)^2 + (y - k)^2 = a^2$.
Comparing the given equation $x^2 - y^2 - 2x + 4y - 3 = 0$ with
$ax^2 + by^2+ 2hxy + 2gx + 2fy + c = 0$, we get:
$a = 1, b = -1, h = 0, g = -1, f = 2, c = -3$
Intersection point $\Big(\frac{\text{hf}-\text{bg}}{\text{ab}-\text{h}^2},\ \frac{\text{gh}-\text{af}}{\text{ab}-\text{h}^2}\Big)=\Big(\frac{-1}{-1},\ \frac{-2}{-1}\Big)=(1,\ 2)$
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:
$(x - 1)^2 + (y - 2)^2 = 1$
$⇒ 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 $x^2 + y^2 + x - 2y - 14 = 0$ 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.

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

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

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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
$ae^2$
✓
$2ae^2$

Answer

Correct option: D.

$2ae^2$

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 $x^2+ y^2+ 2gx + 2fy + c = 0$ does not intersect $x-$axis, if:

✓
$g^2 < c$
B
$g^2 > c$
C
$g^2 > 2c$
D
None of these

Answer

Correct option: A.

$g^2 < c$

Given:
$x^2+ y^2+ 2gx + 2fy + c = 0 ......... (1)$
The given circle intersects the $x-$axis.
The equation of circle becomes $x^2+ 2gx + c = 0 ......... (2)$
Solving equation $(2):$
$\therefore$ Discriminant, $\text{D}=\sqrt{4\text{g}^2-4\text{c}}\geq0$
$\Rightarrow4\text{g}^2-4\text{c}\geq0$
$\Rightarrow\text{g}^2-\text{c}\geq0$
$\Rightarrow\text{g}^2\geq\text{c}$
Hence, if $g^2 < c$, then the given circle will not intersect the $x-$axis.

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MCQ 231 Mark

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

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

Answer

Correct option: B.

$(a, -a)$

Given:
The equation of the parabola is $(y + a)^2 = 8a (x - a)$.
Putting $X = x - a, Y = y + a$
$Y^2= 8aX$
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.

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

Answer

Correct option: C.

$y^2= -8x$

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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.

$2$

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 $y^2 + 6y + 2x + 5 = 0,$ then $SV =$

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

Answer

Correct option: B.

$\frac{1}{2}$

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)b $
Let $Y = y + 3, X = x - 2$
Then, the equation of parabola becomes $Y^2= -2X$.
Vertex $= (X = 0, Y = 0) = (x - 2 = 0, y + 3 = 0) = (x = 2, y = -3)$
Comparing with $y^2= 4ax:$
$4\text{a} = 2$
$\Rightarrow \text{a} =\frac{1}{2}$
Focus $=\ \Big(\text{X}=\frac{-1}{2}, \text{Y}=0\Big)=\Big(\text{x}-2=\frac{-1}{2},\text{y}+3=0\Big)=\Big(\text{x}=\frac{3}{2},\text{y}=-3\Big)$
$\Rightarrow\ \text{SV}=\sqrt{\Big(2-\frac{3}{2}\Big)^2+(-3+3)^2}=\frac{1}{2}$

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MCQ 291 Mark

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

A
$2$
B
$4$
✓
$8$
D
$16$

Answer

Correct option: C.

$8$

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

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MCQ 301 Mark

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

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

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

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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.

$\sqrt{5}$

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.

$1$

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.

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

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 $\text{X = 4aY}$
$\text{a}=\frac{1}{8}$
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 $

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 $(\pm3,0)$ and foci at $(\pm5,0),$ 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$

The vertices of the hyperbola are $(\pm3,0)$ and foci are $(\pm5,0).$
Thus, the value of a and ae are $3$ and $5$, respectively.
Now, using the relation $b^2 = a^2(e^2 - 1)$, we get:
$b^2= 25 - 9$
$\Rightarrow b^2= 16$
Equation of hyperbola is given below:
$\frac{\text{x}^2}{9}-\frac{\text{y}^2}{16}=1$
$ 16 x^2-9 y^2=144$

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MCQ 361 Mark

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

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

Answer

Correct option: A.

$(-4, 1)$

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 $X=x+4, Y=y-1$
$X^2=-12 Y$
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.

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

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MCQ 381 Mark

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

A
$1$
B
$2$
C
$−1$
✓
$4$

Answer

Correct option: D.

$4$

$9x^2- kx^2 + y^2 + ky^2+ 2kx = 0$
$x^2(9 - k) + y^2(1 + k) + 2kx = 0$
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 $

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 $r = 2$
$\therefore$ 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.

$\frac{4}{3}$

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.

$15$

The centre of the circle $x^2+y^2=9$ is $(0,0)$
Let us denote it by $\mathrm{C}_1$.
The centre of the circle $x^2+y^2+8 y+c=0$ is $(0,-4)$.
Let us denote it by $\mathrm{C}_2$.
The radius of $x^2+y^2=9$ is $3$ units.
$x^2+y^2+8 y+c=0$
$\Rightarrow(\text{x}-0)^2+(\text{y}+4)^2=16-\text{c}=(\sqrt{16-\text{c}})^2$
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
$x^2 + y^2 + ax + by = 0$
✓
$x^2 + y^2 - ax - by = 0$
C
$x^2 + y^2 + bx + ay = 0$
D
None of these

Answer

Correct option: B.

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

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.

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

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.

$(-5, 10)$

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.

$(1, 2)$

Given:
$(y - 2)^2 = 16 (x - 1)$
Let $X = x - 1, Y = y - 2$
$\therefore$ $Y^2= 16X$
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.

$16$

$\sqrt{\Big(\frac{-\lambda}{2}^2\Big)+\Big(\frac{\lambda-1}{2}\Big)^2-5}\leq5$
$\Rightarrow\Big(\frac{-\lambda}{2}^2\Big)+\Big(\frac{\lambda-1}{2}\Big)\leq30$
$\lambda^2+(\lambda-1)^2\leq120$
$\Rightarrow2\lambda^2-2\lambda-199\leq0$
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.

$9$

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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.

$4\sqrt2\text{a}$

Let $OP$ be the chord.
Let the coordinates of $P$ be $(x_1, y_1).$
From the figure, we have:
$OP^1 = {x_1}^2+ {y_1}^2...(1)$
And $,\tan\frac{\pi}{4}=\frac{\text{y}_1}{\text{x}_1}$
$\Rightarrow x_1=y_1...(2)$
Also, $(x_1, y_1)$ lies on the parabola.
$\therefore {y_1}^2= 4ax_1...(3)$
Using $(2)$ and $(3):$
${x_1}^2 = 4ax_1$
$\Rightarrow x_1 = 4a ...(4)$
From $(4), (1)$ and $(2),$ we have:
$OP^2= (4a)^2+ (4a)^2= 32a^2$
$\Rightarrow\ \text{OP}=4\sqrt2\text{a}$
Therefore, the length of the chord is $4\sqrt2\text{a}$ 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 $

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

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

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.

$x + y = 5, y = 3$

$\Rightarrow\frac{\text{x}^2}{16}+\frac{\text{y}^2}{9}=1$
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 $

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

$225\pi$

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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.

$2x + y = 0$

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:
$y = -2x$
$\Rightarrow 2x + y = 0$
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.

$2$

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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.

$\pm4$

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

Answer

Correct option: B.

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

The equation of a hyperbola with foci on the $x-$axis is defined as. $\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2} = 1$

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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

$ 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 $

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

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

Answer

Correct option: D.

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

Solution: (D) 2xy - 4x + 4y + 1 = 0
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^2-2 x+1+y^2+1+2 y=x^2+y^2+1-2 x y-2 y+2 x $
$ 2 x y-4 x+4 y+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
$x^2+ y^2± 10x ± 10y + 5 = 0$
B
$x^2+ y^2 ± 10x ± 10y = 0$
✓
$x^2+ y^2± 10x ± 10y + 25 = 0$
D
$x^2+ y^2 ± 10x ± 10y + 51 = 0$

Answer

Correct option: C.

$x^2+ y^2± 10x ± 10y + 25 = 0$

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 $x^2+ y^2- 2ax - 2ay + a^2 = 0.$
The given radius of the circle is $5$ units, i.e. $a = 5.$
Thus, the equation of the circle is $x^2+ y^2 - 10x - 10y + 25 = 0.$
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 $x^2+ y^2 + 2ax - 2ay + a^2= 0$.
The given radius of the circle is $5$ units, i.e. $a = 5.$
Thus, the equation of the circle is $x^2+ y^2 + 10x - 10y + 25 = 0.$
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 $x^2+ y^2 + 2ax + 2ay + a^2= 0$
The given radius of the circle is $5$ units, i.e. $a = 5.$
Thus, the equation of the circle is $x^2+ y^2+ 10x + 10y + 25 = 0.$
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 $x^2+ y^2 - 2ax + 2ay + a^2= 0$.
The given radius of the circle is $5$ units, i.e. $a = 5.$
Thus, the equation of the circle is $x^2+ y^2 - 10x + 10y + 25 = 0.$
Hence, the required equation of the circle is $x^2+ y^2± 10x ± 10y + 25 = 0.$

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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.

$7$

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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.

$(-1,\ 3)$

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 $4x^2+ 4y^2- 8x + 12y - 25 = 0$ is:

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

Answer

Correct option: A.

$(2, -3)$

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MCQ 721 Mark

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

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

Answer

Correct option: A.

$x^2+ y^2 = 64$

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.

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

Let $\text{ABC}$ 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.

$x^2- y^2 = 32$

The distance between the foci is $2ae.$
$\therefore 2ae = 16$
$\Rightarrow ae = 8$
$\text{e}=\sqrt2$
$\therefore\text{a}\sqrt2=8$
$\Rightarrow\text{a}=4\sqrt2$
Also $ b^2=a^2\left(e^2-1\right) $
$\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 $x^2 = 4ay$?

A
$x = at^2, y = 2at$
B
$x = 2at, y = at^2$
C
$x = 2at^2, y = at$
✓
$x = 2at, y = at^2$

Answer

Correct option: D.

$x = 2at, y = at^2$

Substituting $x = 2at, y = at^2$ in the given equation:
$(2at)^2= 4a(at^2)$
$⇒ 4a^2t^2 = 4a^2t^2$
Hence, $(2at, at^2)$ lies on the parabola $x^2 = 4ay$.

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MCQ 761 Mark

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

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

Answer

Correct option: B.

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

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 $

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:

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 $

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

$\text{AG : GD} = 2 : 1$
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=4 a^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 $

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

$(x + 1)^2+ (y - 1)^2 = 2^2$ 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 $3x + 4y - 5 = 0$ is:

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

Answer

Correct option: D.

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

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

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

Answer

Correct option: C.

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

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^2-18 a+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^2-6 x-6 y+9=0$

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MCQ 841 Mark

If $(-3, 2)$ lies on the circle $x^2+ y^2+ 2gx + 2fy + c = 0$ which is concentric with the circle $x^2+ y^2+ 6x + 8y - 5 = 0$, then $c =$

A
$11$
✓
$-11$
C
$24$
D
None of these

Answer

Correct option: B.

$-11$

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:

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

Answer

Correct option: A.

$x^2= -12y$

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) $

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 $

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 $\frac{2\text{b}^2}{\text{a}}$ 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 $9x^2+ 4y^2 = 36$ and $e_2$ is the eccentricity of the conic $9x^2- 4y^2 = 36$, 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.

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

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.

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

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:
$\frac{\text{a}^2}{2}=\text{a}^2-\text{c}$
$\Rightarrow\frac{\text{a}^2}{2}=\text{c}$
$\Rightarrow\frac{2}{\text{a}^2}=\frac{1}{\text{c}}$
$\Rightarrow\frac{1}{\text{a}^2}+\frac{1}{\text{a}^2}=\frac{1}{\text{c}}$

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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.

$12\text{x}^2+4\text{y}^2+24\text{x}-16\text{y}+24=0$
$\Rightarrow12\big(\text{x}^2+2\text{x}\big)+4\big(\text{y}^2-4\text{y}\big)=-24$
$\Rightarrow12\big(\text{x}^2+2\text{x}+1\big)+4\big(\text{y}^2-4\text{y}+4\big)=-24+12+16$
$\Rightarrow12\big(\text{x}+1\big)^2+4\big(\text{y}-2\big)^2=4$
$\Rightarrow\frac{(\text{x}+1)^2}{3}+\frac{(\text{y}-2)^2}{1}=1$
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 $x = t^2 + 1, y = 2t + 1$. The cartesian equation of its directrix is

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

Answer

Correct option: A.

$x = 0$

Given:
$x = t^2+ 1 ...(1)$
$y = 2t + 1 ...(2)$
From $(1)$ and $(2)$:
$\text{x}=\Big(\frac{\text{y}-1}{2}\Big)^2+1$
On simplifying:
$(y - 1)^2= 4(x - 1)$
Let $Y = y - 1$ and $X = x - 1$
$\therefore$ $Y^2= 4X$
Comparing it with $y^2= 4ax:$
$a = 1$
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.

$\frac{4}{3}$

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.

$(2, 0)$

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:
$(\text{y}-2)=\frac{-1}{2}(\text{x}-0)$
$\Rightarrow x + 2y = 4$

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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.

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

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MCQ 971 Mark

If the eccentricity of the hyperbola $x^2 − y^2\ \sec^2 \alpha = 5$ is $\sqrt3$ 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.

$\frac{\pi}{4}$

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$ and $\text{b}^2 = 5\cos^2\alpha.$
$\Rightarrow\text{b}^2 = \text{a}^2(\text{e}_1^2 − 1)$
$\Rightarrow 5\cos^2\alpha=5(\text{e}_1^2−1)$
$\Rightarrow\text{e}_1^2=\cos^2\alpha+1...(1)$
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 $\text{a}^2=25$ 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 981 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 $

Equation of circle which touches the $y-$axis at origin is $x^2+ y^2 + 2gx + d = 0$
Since the circle passes through origin,
$d = 0$ Thus the equation becomes, $x^2+ y^2 + 2gx = 0 ...... (1)$
The equation passes through $(b, c)$
so, $b^2+ c^2+ 2gb = 0$
$\therefore\text{g}=\frac{-\text{b}^2-\text{c}^2}{\text{2b}}$
So, putting the value of $g$ in $(1)$ we get $bx^2+ by^2- (b^2+ c^2) x = 0$

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MCQ 991 Mark

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

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

Answer

Correct option: C.

$y^2= -8x$

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MCQ 1001 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.

$1$

Given circle equation: $x^2+ y^2= 5$
$x^2+ y^2- 5 = 0 … (1)$
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 1011 Mark

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

✓
$(1, 1)$
B
$(1, 0)$
C
$(1, -1)$
D
$(0, 0)$

Answer

Correct option: A.

$(1, 1)$

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

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MCQ 1021 Mark

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

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MCQ 1031 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.

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

To find the centre:
Coefficient of $x^2 =$ Coefficient of $y^2$
$\therefore\lambda=2\lambda-3$
$\Rightarrow\lambda=3$
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 1041 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 1051 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.

$8$

$ 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 $y^2=4 a x$ :
$a = 2$
Length of the latus rectum $= 4a = 8$ units

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MCQ 1061 Mark

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

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

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

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MCQ 1071 Mark

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

A
$x = -a$
B
$x = a$
✓
$y = -a$
D
$y = a$

Answer

Correct option: C.

$y = -a$

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

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MCQ 1081 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) $

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 1091 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 1101 Mark

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

✓
$1$
B
$-1$
C
$21$
D
$16$

Answer

Correct option: A.

$1$

$x^2+ y^2 = a ........ (1)$
And, $x^2+ y^2− 6x − 8y + 9 = 0 ........ (2)$
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 $x^2+ y^2− 6x − 8y + 9 = 0, C_1,$ 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 1111 Mark

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

A
$4$
B
$6$
✓
$8$
D
$10$

Answer

Correct option: C.

$8$

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)$
$ \text { Let } X=x-2, Y=y-1 $
$ \therefore X^2=8 Y$
$\therefore$ Length of the latus rectum $= 4a = 8$ units

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MCQ 1121 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 1131 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.

$\frac{4}{3}$

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 1141 Mark

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

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

Answer

Correct option: B.

A parabola

Comparing the given equation with $ax^2 + by^2 + 2hxy + 2gx + 2fy + c = 0$, we get:
$a = 16, b = 1, h = 4$
We have: $h^2 = 16 = ab$
Thus, the given equation represents a parabola.

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MCQ 1151 Mark

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

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

Answer

Correct option: C.

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

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MCQ 1161 Mark

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

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

Answer

Correct option: A.

$(2, 4)$

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

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MCQ 1171 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 1181 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.

$(\pm5,0)$

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 1191 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 1201 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 $

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MCQ 1211 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 $

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 1221 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 1231 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 $

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 1241 Mark

The locus of the points of trisection of the double ordinates of a parabola is a

A
Pair of lines
B
Circle
✓
Parabola
D
Parabola

Answer

Correct option: C.

Parabola

Suppose $PQ$ is a double ordinate of the parabola $y^2= 4ax$.
Let $R$ and $S$ be the points of trisection of the double ordinates.
Let $(h, k)$ be the coordinates of $R.$
Then, we have:
$OL = h$ and $RL = k$
$\therefore \ce{RS = RL + LS = k + k = 2k}$
$\Rightarrow \ce{PR = RS = SQ = 2K}$
$\Rightarrow \ce{LP = LR + RP = k + 2k = 3k}$
Thus, the coordinates of $P$ are $(h, 3k)$ which lie on $y^2 = 4ax.$
$\therefore 9k^2= 4ah$
Hence, the locus of the point $(h, k)$ is $9\text{y} = 4\text{ax}$
i.e. $\text{y}^2=\Big(\frac{4\text{a}}{9}\Big)\text{x}$ which represents a parabola.

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MCQ 1251 Mark

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

✓
$(2, 0)$
B
$(-2, 0)$
C
$(0, 2)$
D
$(0, -2)$

Answer

Correct option: A.

$(2, 0)$

Given, $y^2= 8x$
General equation is $y^2= 4ax$
Now, $4a = 8$
$\Rightarrow a = 2$
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 $

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

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MCQ 1271 Mark

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

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

Answer

Correct option: B.

$(-1, 2)$

We have,
$y^2- 4y - x + 3 = 0$
$(y - 2)^2- 4 - x + 3 = 0$
$(y - 2)^2= (x + 1)$
$\therefore$ 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.

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

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

$(3, -4), 5$

$ (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 $y = mx + c$ 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.

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

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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.

$\frac{8}{3}$

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 $x^2- 4x - 8y + 12 = 0$ is

A
$y = 0$
B
$x = 1$
✓
$y = -1$
D
$x = -1$

Answer

Correct option: C.

$y = -1$

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)$
Putting $X = x - 2, Y = y - 1:$
$X^2= 8Y$
Comparing with $X^2= 4aY:$
$a = 2$
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.

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

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

Answer

Correct option: D.

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

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 $3x - 4y - 7 = 0$ and $2s - 3y - 5 = 0$ 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 $

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MCQ 1381 Mark

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

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

Answer

Correct option: B.

$x + 3y = 0$

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

$3$

$\Rightarrow3(\text{x}^2-2\text{x})+4(\text{y}^2+2\text{y})=5$
$\Rightarrow3(\text{x}^2-2\text{x}+1)+4(\text{y}^2+2\text{y}+1)=5+3+4$
$\Rightarrow3(\text{x}-1)^2+4(\text{y}+1)^2=12$
$\frac{(\text{x-1})^2}{4}+\frac{(\text{y}+1)^2}{3}=1$
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.

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

Let the equation of the circle be,
$(x-h)^2+(y-k)^2=r^2$
Let the centre be $(0, a)$
$\therefore$ Radius $\mathrm{r}=\mathrm{a}$
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^2-2 a y=a^2 $
$ \Rightarrow x^2+y^2-2 a y=0 \ldots .(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 $(4a - 3) x^2+ ay^2+ 6x - 2y + 2 = 0$ represents a circle, then its centre is:

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

Answer

Correct option: C.

$(-3, 1)$

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

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MCQ 1431 Mark

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

A
$2a$
B
$3a$
C
$6a$
✓
$4a$

Answer

Correct option: D.

$4a$

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

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MCQ 1441 Mark

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

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

Answer

Correct option: A.

$\frac{5}{4}$

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

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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.

$x^2+y^2=6400$

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

Answer

Correct option: A.

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

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:
$\ce{SP = PM}$
$\Rightarrow \ce{SP^2= PM^2}$
$\Rightarrow\ (\text{x}-0)^2+(\text{y}-0)^2=\Big(\frac{\text{x+y}-4}{\sqrt2}\Big)^2$
$\Rightarrow\ \text{x}^2+\text{y}^2=\Big(\frac{\text{x+y}-4}{\sqrt2}\Big)^2$
$\Rightarrow 2x^2+ 2y^2= x^2+ y^2+ 16 + 2xy - 8x - 8y$
$ \Rightarrow x^2+y^2-2 x y+8 x+8 y-16=0 $

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MCQ 1471 Mark

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

✓
$3$
B
$6$
C
$\frac{1}{2}$
D
$9$

Answer

Correct option: A.

$3$

Given:
$4y^2+ 2x - 20y + 17 = 0$
$\Rightarrow\ \text{y}^2+\frac{\text{x}}{2}-5\text{y}+\frac{17}{4}=0$
$\Rightarrow\ \Big(\text{y}-\frac{5}{2}\Big)^2+\frac{\text{x}}{2}-2=0$
$\Rightarrow\ \Big(\text{y}-\frac{5}{2}\Big)^2=-1\Big(\frac{\text{x}}{2}-2\Big)$
$\Rightarrow\ \Big(\text{y}-\frac{5}{2}\Big)^2=\frac{-1}{2}(\text{x}-4)$
$\text{Let }\text{X}=\text{x}-4,\ \text{Y}=\text{y}-\frac{5}{2}$
$\therefore\ \text{Y}^2=\frac{-\text{X}}{2}$
$\therefore$ 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.

$10$

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

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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.

$154$

The equation $x^2+ y^2= 49$ 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 $3x^2+ y^2= 12$ is:

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

Answer

Correct option: D.

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

$3\text{x}^2+\text{y}^2=12$
$\Rightarrow\frac{\text{x}^2}{4}+\frac{\text{y}^2}{12}=1$
$\therefore\text{ a}^2=4$
$\Rightarrow\text{a}=2$
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 $\ce{P(x,y)}$ be any point on the hyperbola, and $\ce{S, S'}$ be the focus with coordinates $(\pm\text{ae},0).$
$\Rightarrow \ce{S'P − SP = 2a}$
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 $PA − PB = k (k \neq 0),$ 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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