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
$-\frac{3}{4}$
AnswerCorrect 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 $(a = 6)$
$\frac{\lambda}{3}=\frac{1}{4}$
$\lambda=\frac{3}{4}$
View full question & answer→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
$\text{none of these}$
AnswerCorrect 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:
$\text{e}^2\text{e}-1=0$
$\Rightarrow\text{e}=\frac{-1\pm\sqrt{1+4}}{2}$
$\Rightarrow\text{e}=\frac{\sqrt{5}-1}{2}$ $(\because\text{e}\text{ cannot be negative}\big)$
View full question & answer→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
AnswerCorrect option: B. $ x^2+y^2-6 x+4 y=3 $
View full question & answer→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
$\frac{\text{x}^2}{9}+\frac{\text{y}^2}{25}=1$
AnswerCorrect option: B. $\frac{\text{x}^2}{25}+\frac{\text{y}^2}{9}=1$
View full question & answer→MCQ 51 Mark
The equation of the incircle formed by the coordinate axes and the line $4x + 3y = 6$ is:
AnswerCorrect 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:
$\Big(\text{x}-\frac{1}{2}\Big)^2+\Big(\text{y}-\frac{1}{2}\Big)^2=\frac{1}{4}$
$\Rightarrow4(\text{x}^2+\text{y}^2)-\text{x}-\text{y}+1=0$
View full question & answer→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
AnswerCorrect 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 $ View full question & answer→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
$(-1, -2)$
AnswerCorrect 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, $\text{x}_1=\frac{\text{at}_1^2+\text{at}_2^2}{2}=\text{t}_1^2+\text{t}_2^2=-1$
$\Rightarrow\ (\text{x}_1, \text{y}_1)=(-1, 2)$
View full question & answer→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
$(1, 0), (-1, 0)$
- ✓
Answer$x^2+ y^2- 3x - 4y + 2 = 0$
$x-$axis will be cut when $y = 0$
put $y=0$
$x^2 - 3x + 2 = 0$
$(x - 2) (x - 1) = 0$
$x = 1, 2$
points $(1, 0), (2, 0)$
View full question & answer→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
AnswerCorrect option: A. $x^2+y^2-4 x+2 y=45$
View full question & answer→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
AnswerCorrect 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 $
View full question & answer→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
$\frac{1}{5}$
AnswerCorrect 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, $\text{e}=\sqrt{1-\frac{\text{b}^2}{\text{a}^2}}$
$\Rightarrow\text{e}=\sqrt{1-\frac{9}{25}}$
$\Rightarrow\text{e}=\sqrt{\frac{16}{25}}$
$\Rightarrow\text{e}=\frac{4}{5}$
View full question & answer→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
AnswerCorrect 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:
$\text{SP = PM}$
$\Rightarrow \ce{SP^2 = PM^2}$
$\Rightarrow\ (\text{x}-0)^2+(\text{y+a})^2=\Big(\frac{\text{x+y}-3\text{a}}{\sqrt2}\Big)^2$
$\Rightarrow\ \text{x}^2+(\text{y+a})^2=\Big(\frac{\text{x+y}-3\text{a}}{\sqrt2}\Big)^2$
$\Rightarrow\ 2\text{x}^2+2\text{y}^2+2\text{a}^2+4\text{ay}=\text{x}^2+\text{y}^2+9\text{a}^2+2\text{xy}-6\text{ax}-6\text{ay}$
$\Rightarrow\ \text{x}^2+\text{y}^2-7\text{a}^2+10\text{ay}+6\text{ax}=0$ View full question & answer→MCQ 131 Mark
The equation $\sqrt{(\text{x}-2)^{2}+\text{y}^{2}}+\sqrt{(\text{x}+2)^{2}+\text{y}^{2}}=5$ represents:
Answerlet $A (2, 0), B (-2, 0)$ 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.
View full question & answer→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:
AnswerThe 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:
$\Bigg|\frac{4\text{c}+3\text{c}-12}{\sqrt{4^2+3^3}}\Bigg|=\text{c}$
$\Rightarrow\frac{7\text{c}-12}{5}=\text{c}$
$\Rightarrow\text{c}=6$ View full question & answer→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
AnswerCorrect option: B. $ x^2+y^2-x-y=0 $
View full question & answer→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
AnswerCorrect 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
$\frac{\text{a}^2\text{e}^2}{4}=\text{a}^2(\text{e}^2-1)$
$\Rightarrow\text{e}^2=4\text{e}^2-4$
$\Rightarrow\text{e}^2=\frac{4}{3}$
$\Rightarrow\text{e}=\frac{2}{\sqrt{3}}$
View full question & answer→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
AnswerCorrect 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 $
View full question & answer→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
AnswerCorrect 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 $
View full question & answer→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
AnswerCorrect 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 $
View full question & answer→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
$(-\infty,\ -2)\cup(3,\ \infty)$
AnswerCorrect 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 $k < -3$
$\Rightarrow\text{k}\in(-\infty,\ -3)\cup(4,\ \infty)$
View full question & answer→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
$ae^2$
- ✓
$2ae^2$
AnswerCorrect 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 $=2\text{a}-2\text{a}(1-\text{e}^2)$
$\\=2\text{a}-2\text{a}+2\text{ae}^2\\=2\text{ae}^2$
View full question & answer→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
AnswerCorrect 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.
View full question & answer→MCQ 231 Mark
The vertex of the parabola $(y + a)^2 = 8a (x - a)$ is
- A
$(-a, -a)$
- ✓
$(a, -a)$
- C
$(-a, a)$
- D
AnswerCorrect 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 $(a, -a).$
View full question & answer→MCQ 241 Mark
The equation $x^2 + y^2 + 2x - 4y + 5 = 0$ represents:
- ✓
- B
A pair of straight lines.
- C
A circle of non-zero radius..
- D
AnswerThe 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.
View full question & answer→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
$y^2= 8x$
AnswerCorrect option: C. $y^2= -8x$
View full question & answer→MCQ 261 Mark
What is the approximate radius of the circle whose equation is $(\text{x}-\sqrt{3})^2+(\text{y}+2)^2=11$:
AnswerCorrect option: C. $3.32$
The radius of given circle is $\sqrt{11}=3.32$
View full question & answer→MCQ 271 Mark
Find the value of a if $y^2= 4ax$ pases through $(8, 8):$
AnswerGiven point $(8, 8)$
Given equation $y^2= 4ax$
$\Rightarrow8^2=4\text{a}(8)$
$64 = 32\text{a}$
$\text{a}=\frac{64}{32}$
$\text{a} = 2$
View full question & answer→MCQ 281 Mark
If $V$ and $S$ are respectively the vertex and focus of the parabola $y^2 + 6y + 2x + 5 = 0,$ then $SV =$
AnswerCorrect 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 $=\ \Big(\text{X}=\frac{-1}{2}, \text{Y}=0\Big)=\Big(\text{x}-2=\frac{-1}{2},\text{y}+3=0\Big)=\Big(\text{x}=\frac{3}{2},\text{y}=-3\Big)$
$\Rightarrow\ \text{SV}=\sqrt{\Big(2-\frac{3}{2}\Big)^2+(-3+3)^2}=\frac{1}{2}$
View full question & answer→MCQ 291 Mark
The length of latus rectum of the parabola $y^2 + 8x - 2y + 17 = 0$ is:
AnswerThe 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 $= 4a = 4 \times 2 = 8$
View full question & answer→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
AnswerCorrect 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 $\Big(\frac{5}{4}, 1\Big)$
View full question & answer→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
$\sqrt{7}$
AnswerCorrect 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 $
$\text{c}=\sqrt{5}$
View full question & answer→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 $2$
AnswerGiven, 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.
View full question & answer→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
$\Big(-\frac{1}{4}, \frac{1}{8}\Big)$
AnswerCorrect 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 $\Big(-\frac{1}{4},0\Big).$
View full question & answer→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
AnswerCorrect 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 $
View full question & answer→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 $
AnswerCorrect 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$
View full question & answer→MCQ 361 Mark
The vertex of the parabola $x^2 + 8x + 12y + 4 = 0$ is
- ✓
$(-4, 1)$
- B
$(4, -1)$
- C
$(-4, -1)$
- D
$(4, 1)$
AnswerCorrect 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 $(-4, 1).$
View full question & answer→MCQ 371 Mark
The equation of the circle $x^2+ y^2+ 2gx + 2fy + c = 0$ will represent a real circle if:
AnswerCorrect option: B. $\text{g}^{2} + \text{f}^{2} – \text{c} \underline{>} 0$
View full question & answer→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?
Answer$9x^2- kx^2 + y^2 + ky^2+ 2kx = 0$
$x^2(9 - k) + y^2(1 + k) + 2kx = 0$
for circle
$9 - k = 1 + k$
So, $k = 4$
View full question & answer→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
AnswerCorrect 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 $
View full question & answer→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
$4$
AnswerCorrect 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 $= \text{4a} = 4 \times (\frac{1}{3}) = \frac{4}{3}$
View full question & answer→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:
AnswerThe 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 $P.$
$\therefore\text{C}_1\text{C}_2=\text{PC}_2+\text{PC}_1$
$\Rightarrow\text{PC}_2=4-3=1$
$\Rightarrow\text{PC}_2-1=\sqrt{16-\text{c}}$
$\Rightarrow\text{c}=15$
View full question & answer→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
AnswerCorrect 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:
$\Big(\text{x}-\frac{\text{a}}{2}\Big)^2+\Big(\text{y}-\frac{\text{b}}{2}\Big)^2=\frac{1}{4}(\text{a}^2+\text{b}^2)$
$\Rightarrow(2\text{x}-\text{a}^2)+(2\text{y}-\text{b})^2=(\text{a}^2+\text{b}^2)$
$\Rightarrow4\text{x}^2+\text{a}^2-4\text{ax}+4\text{y}^2+\text{b}^2-4\text{by}=\text{a}^2+\text{b}^2$
$\Rightarrow\text{x}^2-\text{ax}+\text{y}^2-\text{by}=0$
View full question & answer→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
$\text{e}_1^2 + 3\text{e}_2^2 = 2$
AnswerCorrect 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:
$\text{b}^2 = \text{a}^2(\text{e}_2^2 - 1)$
$\Rightarrow4 = 9(\text{e}_2^2 - 1)$
$\Rightarrow\frac{4}{9}=\text{e}_2^2-1$
$\Rightarrow\text{e}_2^2=\frac{13}{9}$
$\therefore2\text{e}_1^2+\text{e}_2^2=\frac{2\times7}{9}+\frac{13}{9}$
$=\frac{27}{9}$
$=3$
View full question & answer→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
AnswerCorrect 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 $(−1, 2).$
View full question & answer→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
AnswerCorrect 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 $(-5, 10)$
View full question & answer→MCQ 461 Mark
The vertex of the parabola $(y - 2)^2= 16 (x - 1)$ is
- ✓
$(1, 2)$
- B
$(-1, 2)$
- C
$(1, -2)$
- D
$(2, 1)$
AnswerCorrect 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 $(1, 2).$
View full question & answer→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:
Answer$\sqrt{\Big(\frac{-\lambda}{2}^2\Big)+\Big(\frac{\lambda-1}{2}\Big)^2-5}\leq5$
$\Rightarrow\Big(\frac{-\lambda}{2}^2\Big)+\Big(\frac{\lambda-1}{2}\Big)\leq30$
$\lambda^2+(\lambda-1)^2\leq120$
$\Rightarrow2\lambda^2-2\lambda-199\leq0$
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 $16.$
View full question & answer→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 $\lambda=$
View full question & answer→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
AnswerCorrect 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. View full question & answer→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
AnswerCorrect 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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