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. $\frac{3}{4}$
- $\frac{3}{4}$
Solution:
$\frac{\lambda(\text{x}+1)^2}{3}+\frac{(\text{y}+2)^2}{4}=1$
for a circle of a $(\text{x}-\alpha)^2+6 (\text{y}-\beta)^2=1$ then (a = 6)
$\frac{\lambda}{3}=\frac{1}{4}$
$\lambda=\frac{3}{4}$ 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
$x^2+y^2+3 x-2 y=-3$
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. $\frac{\text{x}^2}{25}+\frac{\text{y}^2}{9}=1$
- $\frac{\text{x}^2}{25}+\frac{\text{y}^2}{9}=1$
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. $4\left(x^2+y^2-x-y\right)+1=0$
- $4\left(x^2+y^2-x-y\right)+1=0$
Solution:
The line 4x + 3y = 6 cuts the coordinate axes at $\Big(\frac{3}{2},\ 0\Big)$ and (0, 2)
The coordinates of the incentre is $\Big(\frac{\text{ax}_1+\text{bx}_2+\text{cx}_3}{\text{a+b+c}},\ \frac{\text{ay}_1+\text{by}_2+\text{cy}}{\text{a+b+c}}\Big)$
Here, $\text{a}=\frac{5}{2},\ \text{b}=\frac{3}{2},\ \text{c}=2,\ \text{x}_1=0,\ \text{y}_1\\=0,\ \text{x}_2=0,\ \text{y}_2=2,\ \text{x}_3=\frac{3}{2},\ \text{y}_3=0$
Thus, the coordinates of the incentre:
$\Big(\frac{0+0+3}{6},\ \frac{0+3+0}{6}\Big)$
$=\big(\frac{1}{2},\ \frac{1}{2}\Big)$
The equation of the incircle:
$\Big(\text{x}-\frac{1}{2}\Big)^2+\Big(\text{y}-\frac{1}{2}\Big)^2=\text{a}^2$
Also, radius of the incircle $=\frac{\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}}{\text{s}}$
Here, $\text{s}=\frac{\text{a+b+c}}{2}=\frac{\frac{5}{2}+\frac{3}{2}+2}{2}=3$
$\therefore$ Radius of the incircle $=\sqrt{\frac{3(3-\text{a})(3-\text{b}(3-\text{c}))}{3}}$
$=\frac{\sqrt{3\Big(3-\frac{5}{2}\Big)\Big(3-\frac{3}{2}\Big)(3-\text{c})}}{3}$
$=\frac{\sqrt{3\Big(3-\frac{1}{2}\Big)\Big(\frac{3}{2}\Big)}}{3}$
$=\frac{1}{2}$
The equation of circle:
$\Big(\text{x}-\frac{1}{2}\Big)^2+\Big(\text{y}-\frac{1}{2}\Big)^2=\frac{1}{4}$
$\Rightarrow4(\text{x}^2+\text{y}^2)-\text{x}-\text{y}+1=0$ 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. $x^2+y^2-2 x-2 y-3=0$
- $x^2+y^2-2 x-2 y-3=0$
Solution:
The centre of the circumcircle is (1, 1).
Radius of the circumcircle
$\therefore$ Equation of the circle: $=\sqrt{(1+1)^2+(1-2)^2}=\sqrt{5}$
$(x-1)^2+(y-1)^2=5$
$\Rightarrow x^2+y 2-2 x-2 y-3=0$ 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
Answer
- (-1, 2)
Solution:
Let the coordinates of P and Q be $\left(a \mathrm{t}_1{ }^2, 2 \mathrm{a} \mathrm{t}_1\right)$ and $\left(\mathrm{at}_2{ }^2, 2 \mathrm{a} \mathrm{t}_2\right)$, respectively.
Slope of $\mathrm{PQ}=\frac{2 \mathrm{at}_2-2 \mathrm{at}}{\mathrm{at}_2^2-\mathrm{at}_1^2} \ldots$(1)
But, the slope of $P Q$ is equal to the slope of $2 x-y+4=0$.
$\therefore$ Slope of $\mathrm{PQ}=\frac{-2}{-1}=2$
From (1),
$\frac{2 \mathrm{at}_2-2 \mathrm{a} t_1}{\mathrm{at}_2^2-\mathrm{at}_1^2}=2 \ldots$
Putting $4 a=8$,
$a=2$
$\therefore$ Focus of the given parabola $=(a, 0)=(2,0)$
Using equation (2):
$\frac{4\left(t_2-t_1\right)}{2\left(t_2^2-t_1^2\right)}=2$
$\frac{\left(t_2-t_2\right)}{\left(t_2^2-t_1^2\right)}=1$
$\Rightarrow t_1+t_2=1$
As, points $P$ and $Q$ lie on $2 x-y+4=0$
$\Rightarrow P\left(a t_1{ }^2, 2 a t_1\right) \text { or } P\left(2 t_1{ }^2, 4 t_1\right) \text { lie on line } 2 x-y+4=0$
$\Rightarrow 2\left(2 t_1{ }^2\right)-\left(4 t_1\right)+4=0$
$\Rightarrow t_1{ }^2-t_1+1=0 \ldots(3)$
Also, $Q\left(a t_2{ }^2, 2 a t_2\right)$ or $P\left(2 t_2{ }^2, 4 t_2\right)$ lie on line $2 x-y+4=0$
$\Rightarrow 2\left(2 t_2^2\right)-\left(4 t_2\right)+4=0$
$\Rightarrow t_2^2-t_2+1=0 \ldots(4)$
Adding (3) and (4), we get,
$\Rightarrow t_1^2-t_1+1+t_2{ }^2-t_2+1=0$
$\Rightarrow\left(t_1^2+t_2^2\right)-\left(t_1+t_2\right)+2=0$
$\Rightarrow\left(t_1^2+t_2^2\right)-1+2=0\left[t_1+t_2=1, \text { proved above }\right]$
$\Rightarrow\left(t_1^2+t_2{ }^2\right)=-1$
Let $\left(x_1, y_1\right)$ be the mid-point of PQ.
Then, we have:
$\mathrm{y}_1=\frac{2 \mathrm{a} t_2+2 \mathrm{a} t_1}{2}=2\left(\mathrm{t}_1+\mathrm{t}_2\right)=2$
$\text { And, } \mathrm{x}_1=\frac{a \mathrm{t}_1^2+a \mathrm{t}_2^2}{2}=\mathrm{t}_1^2+\mathrm{t}_2^2=-1$
$\Rightarrow\left(\mathrm{x}_1, \mathrm{y}_1\right)=(-1,2)$ View full question & answer→MCQ 81 Mark
The circle $x^2+y^2-3 x-4 y+2=0$ cuts x-axis:
Answer
- (1, 0), (2, 0)
Solution:
$x^2+y^2-3 x-4 y+2=0$
$x \text {-axis will be cut when } y=0$
$\text { put } y=0$
$x^2-3 x+2=0$
$(x-2)(x-1)=0$
$x=1,2$
$\text { points }(1,0),(2,0)$ 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
$x^2+y^2-4 x+2 y+45=0$
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
$(x+1)^2+(y-3)^2=25$
AnswerCorrect option: B. $(x-1)^2+(y+3)^2=25$
- $(x-1)^2+(y+3)^2=25$
Solution:
The general equation of a circle with center at $(a, b)$ and radius $r$ is $(x-a)^2+(y-b)^2=r^2$
So substituting the values we get the circle equation as $(x-1)^2+(y+3)^2=25$ 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. $\frac{4}{5}$
$9\text{x}^2+25\text{y}^2=225$
$\Rightarrow\frac{\text{x}^2}{25}+\frac{\text{y}^2}{9}=1$
Comparing it with $\Rightarrow\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}=1,$ we get:
$\text{a}=5$ and $\text{b}=3$
Here, a > b, so the major and the minor axes of the ellipse are along the x−axis and y−axis, respectively.
Now, $\text{e}=\sqrt{1-\frac{\text{b}^2}{\text{a}^2}}$
$\Rightarrow\text{e}=\sqrt{1-\frac{9}{25}}$
$\Rightarrow\text{e}=\sqrt{\frac{16}{25}}$
$\Rightarrow\text{e}=\frac{4}{5}$
View full question & answer→MCQ 121 Mark
The equation of the parabola whose vertex is $(a, 0)$ and the directrix has the equation $x+y=3 a$, is
- A
$x^2+y^2+2 x y+6 a x+10 a y+7 a^2=0$
- ✓
$x^2-2 x y+y^2+6 a x+10 a y-7 a^2=0$
- C
$x^2-2 x y+y^2-6 a x+10 a y-7 a^2=0$
- D
AnswerCorrect option: B. $x^2-2 x y+y^2+6 a x+10 a y-7 a^2=0$
- $x^2-2 x y+y^2+6 a x+10 a y-7 a^2=0$
Solution:
Given:
The vertex is at (a, 0) and the directrix is the line x + y = 3a.
The slope of the line perpendicular to x + y = 3a is 1.
The axis of the parabola is perpendicular to the directrix and passes through the vertex.
$\therefore$ Equation of the axis of the parabola = y − 0 = 1(x - a) ...(1)
Intersection point of the directrix and the axis is the intersection point of (1) and x + y = 3a.
Let the intersection point be K.
Therefore, the coordinates of K are (2a, a)
The vertex is the mid-point of the segment joining K and the focus (h, k).
$\therefore\ \text{a}=\frac{2\text{a+h}}{2},\ 0=\frac{\text{a+k}}{2}$
h = 0, k = -a
Let P (x, y) be any point on the parabola whose focus is S (h, k) and the directrix is x + y= 3a.
Draw PM perpendicular to x + y = 3a.
Then, we have:
$SP = PM$
$\Rightarrow SP^2 = PM^2$
$\Rightarrow\ (\text{x}-0)^2+(\text{y+a})^2=\Big(\frac{\text{x+y}-3\text{a}}{\sqrt2}\Big)^2$
$\Rightarrow\ \text{x}^2+(\text{y+a})^2=\Big(\frac{\text{x+y}-3\text{a}}{\sqrt2}\Big)^2$
$\Rightarrow\ 2\text{x}^2+2\text{y}^2+2\text{a}^2+4\text{ay}=\text{x}^2+\text{y}^2+9\text{a}^2+2\text{xy}-6\text{ax}-6\text{ay}$
$\Rightarrow\ \text{x}^2+\text{y}^2-7\text{a}^2+10\text{ay}+6\text{ax}=0$ 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, $\sqrt{(\text{x}-2)^{2}+\text{y}^{2}}+\sqrt{(\text{x}+2)^{2}+\text{y}^{2}}=5>\text{AB}$
⇒ 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:
Answer
- 6
Solution:
The equation of the circle that touches the axes of coordinates is $x^2 + y^2 - 2cx − 2cy + c^2 = 0$.
Also, $x^2 + y^2 − 2cx − 2cy + c^2 = 0$ touches the line $\frac{\text{x}}{3}+\frac{\text{y}}{4}=1$ or 4x +3y -12 = 0.
Since the circle lies in the first quadrant, it centre is is (c, c).
From the figure, we have:
$\Bigg|\frac{4\text{c}+3\text{c}-12}{\sqrt{4^2+3^3}}\Bigg|=\text{c}$
$\Rightarrow\frac{7\text{c}-12}{5}=\text{c}$
$\Rightarrow\text{c}=6$ View full question & answer→MCQ 151 Mark
The intercept on the line $y=x$ by the circle $x^2+y^2-2 x=0$ is $A B$. Equation of the circle with $A B$ as a diameter is:
- A
$x^2+y^2+x+y=0$
- ✓
$x^2+y^2-x-y=0$
- C
$x^2+y^2+x-y=0$
- D
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
$(x-1)^2+(y+1)^2=25$
AnswerCorrect option: C. $x^2+y^2=25$
- $x^2+y^2=25$
Solution:
Compare the equation with the standard form with center at $(\mathrm{h}, \mathrm{k})$ and radius r is.
$(x-h)^2+(y-k)^2=r^2$
Substitute the value of $(h, k)=(0,0)$ and $r=5$.
Then, the equation becomes $x^2+y^2=25$ 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
$y^2=2 x-4$
- B
$x^2=4 y-8$
- ✓
$y^2=4 x-8$
- D
AnswerCorrect option: C. $y^2=4 x-8$
- $y^2=4 x-8$
Solution:
$y^2=4 a x$
$(y-0)^2=4 a(x-2)$
$y^2=4 a(x-2)$
$y^2=4(x-2)$
$y^2=4 x-8$ 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 $x^2-y^2-$ $2 x+4 y-3=0$, is:
- ✓
$x^2+y^2-2 x-4 y+4=0$
- B
$x^2+y^2+2 x+4 y-4=0$
- C
$x^2+y^2-2 x+4 y+4=0$
- D
AnswerCorrect option: A. $x^2+y^2-2 x-4 y+4=0$
- $x^2+y^2-2 x-4 y+4=0$
Solution:
Let the required equation of the circle be $(x-h)^2+(y-k)^2=a^2$.
Comparing the given equation $x^2-y^2-2 x+4 y-3=0$ with $a x^2+b y^2+2 h x y+2 g x+2 f y+c=0$, we get:
$a=1, b=-1, h=0, g=-1, f=2, c=-3$
Intersection point $\left(\frac{\mathrm{hf}-\mathrm{bg}}{\mathrm{ab}-\mathrm{h}^2}, \frac{\mathrm{gh}-\mathrm{af}}{\mathrm{ab}-\mathrm{h}^2}\right)=\left(\frac{-1}{-1}, \frac{-2}{-1}\right)=(1,2)$
Thus, the centre of the circle is $(1,2)$
The equation of the required circle is $(x-1)^2+(y-2)^2=a^2$
Since circle passes through ( 1,1 ), we have:
$1=a^2$
$\therefore$ Equation of the required circle:
$(x-1)^2+(y-2)^2=1$
$\Rightarrow x^2+y^2-2 x-4 y+4=0$ View full question & answer→MCQ 201 Mark
If the point $(2, k)$ lies outside the circles $x^2+y^2+x-2 y-14=0$ and $x^2+y^2=13$ then klies in the interval:
- A
$(-3,\ -2)\cup(3,\ 4)$
- B
$-3,\ 4$
- ✓
$(-\infty,\ -3)\cup(4,\ \infty)$
- D
$(-\infty,\ -2)\cup(3,\ \infty)$
AnswerCorrect option: C. $(-\infty,\ -3)\cup(4,\ \infty)$
- $(-\infty,\ -3)\cup(4,\ \infty)$
Solution:
The given equations of the circles are $x^2+y^2+x-2 y-14=0$ and $x^2+y^2=13$.
Since $(2, k)$ lies outside the given circles, we have:
$4+k^2+2-2 k-14>0 \text { and } 4+k^2>13$
$\Rightarrow k^2-2 k-8>0 \text { and } k^2>9$
$\Rightarrow(k-4)(k+2)>0 \text { and } k^2>9$
$\Rightarrow k>4 \text { or } k<-2 \text { and } k>3 \text { or } k<-3$
$\Rightarrow k>4 \text { and } k<-3$
$\Rightarrow k \in(-\infty,-3) \cup(4, \infty)$ View full question & answer→MCQ 211 Mark
The difference between the lengths of the major axis and the latus-rectum of an ellipse is
AnswerCorrect option: D. $2 a e^2$
- $2 a e^2$
Solution:
Length of the latus rectum $=\frac{2\text{b}^2}{\text{a}}$
and $\text{e}=\sqrt{1-\frac{\text{b}^2}{\text{a}^2}}$
$\text{a}^2\text{e}^2=\text{a}^2-\text{b}^2$
$\Rightarrow\text{b}^2=\text{a}^2-\text{a}^2\text{e}^2$
$\Rightarrow\text{b}^2=\text{a}^2(1-\text{e}^2)$
$\therefore\ $Length of the latus rectum $=\frac{2\text{a}^2(1-\text{e}^2)}{\text{a}}=2\text{a}(1-\text{e}^2)$
Length of the major axis = 2a
Difference between length of latus rectum and length of major axis $=2\text{a}-2\text{a}(1-\text{e}^2)$
$\\=2\text{a}-2\text{a}+2\text{ae}^2\\=2\text{ae}^2$ View full question & answer→MCQ 221 Mark
The circle $x^2+y^2+2 g x+2 f y+c=0$ does not intersect $x$-axis, if:
AnswerCorrect option: A. $\mathrm{g}^2<\mathrm{c}$
- $\mathrm{g}^2<\mathrm{c}$
Solution:
Given:
$x^2+y^2+2 g x+2 f y+c=0$ ......... (1)
The given circle intersects the $x$-axis.
The equation of circle becomes $x^2+2 \mathrm{gx}+\mathrm{c}=0$ ......... (2)
Solving equation (2):
$\therefore \text { Discriminant, } D=\sqrt{4 g^2-4 c} \geq 0$
$\Rightarrow 4 g^2-4 c \geq 0$
$\Rightarrow g^2-c \geq 0$
$\Rightarrow g^2 \geq c$
Hence, if $\mathrm{g}^2<\mathrm{c}$, then the given circle will not intersect the x -axis. View full question & answer→MCQ 231 Mark
The vertex of the parabola $(y+a)^2=8 a(x-a)$ is
Answer
- (a, a)
Solution:
Given:
The equation of the parabola is $(y+a)^2=8 a(x-a)$.
Putting $X=x-a, Y=y+a$
$Y^2=8 a X$
Vertex $=(X=0, Y=0)=(x-a=0, y+a=0)=(x=a, y=-a)$
Hence, the vertex is at $(a, a)$. 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
Answer
- A point
Solution:
The radius of the given circle $=\sqrt{1^1+(-2)^2-5}=0$
Hence, the radius of the given circle is zero, which represents a point. View full question & answer→MCQ 251 Mark
The equation of parabola with vertex at origin and directrix. x - 2 = 0 is:
- A
$y^2=-4 x$
- B
$y^2=4 x$
- ✓
$y^2=-8 x$
- D
$y^2=8 x$
AnswerCorrect option: C. $y^2=-8 x$
View full question & answer→MCQ 261 Mark
What is the approximate radius of the circle whose equation is $(\text{x}-\sqrt{3})^2+(\text{y}+2)^2=11$:
AnswerThe 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):
Answer
- 2
Solution:
Given point (8, 8)
Given equation $y^2= 4ax$
$\Rightarrow8^2=4\text{a}(8)$
$64 = 32\text{a}$
$\text{a}=\frac{64}{32}$
$\text{a} = 2$ View full question & answer→MCQ 281 Mark
If $V$ and $S$ are respectively the vertex and focus of the parabola $y^2+6 y+2 x+5=0$, then $S V=$
AnswerCorrect option: B. $\frac{1}{2}$
- $\frac{1}{2}$
Solution:
Given:
The vertex and the focus of a parabola are V and S, respectively.
The given equation of parabola can be rewritten as follows:
$(y+3)^2-9+5+2 x=0$
$\Rightarrow(y+3)^2+2 x=4$
$\Rightarrow(y+3)^2=4-2 x$
$\Rightarrow(y+3)^2=-2(x-2)$
$\text { Let } Y=y+3, X=x-2$
Then, the equation of parabola becomes $Y^2=-2 X$.
$\text { Vertex }=(X=0, Y=0)=(x-2=0, y+3=0)=(x=2, y=-3)$
Comparing with $y^2=4 a x$ :
$4 \mathrm{a}=2 \Rightarrow \mathrm{a}=\frac{1}{2}$
$\text { Focus }=\left(\mathrm{X}=\frac{-1}{2}, \mathrm{Y}=0\right)=\left(\mathrm{x}-2=\frac{-1}{2}, \mathrm{y}+3=0\right)=\left(\mathrm{x}=\frac{3}{2}, \mathrm{y}=-3\right)$
$\Rightarrow \mathrm{SV}=\sqrt{\left(2-\frac{3}{2}\right)^2+(-3+3)^2}=\frac{1}{2}$ View full question & answer→MCQ 291 Mark
The length of latus rectum of the parabola $y^2+8 x-2 y+17=0$ is:
Answer
- 8
Solution:
The given parabola is, $y^2+8 x-2 y+17=0$
$\Rightarrow\left(y^2-2 y+1\right)=-8 x-17+1=-8 x-16$
$\Rightarrow(y-1)^2=-8(x+2)$
Comparing with standard parabola $Y^2=-4 a X$
$Y=y-1, X=x+2, a=2$
Hence length of latus rectum is $=4 a=4 \times 2=8$ View full question & answer→MCQ 301 Mark
The coordinates of the focus of the parabola $y^2-x-2 y+2=0$ are
- ✓
$\Big(\frac{5}{4}, 1\Big)$
- B
$\Big(\frac{1}{4}, 0\Big)$
- C
$(1, 1)$
- D
AnswerCorrect option: A. $\Big(\frac{5}{4}, 1\Big)$
- $\Big(\frac{5}{4}, 1\Big)$
Solution:
Given:
The equation of the parabola is $y^2-x-2 y+2=0$.
$\Rightarrow(\mathrm{y}-1)-1=(\mathrm{x}-2)$
$(\mathrm{y}-1)=\mathrm{x}-1$
$\text { Let } \mathrm{X}=\mathrm{x}-1, \mathrm{Y}=\mathrm{y}-1$
$\mathrm{Y}=\mathrm{X}$
Comparing with $\mathrm{Y}=4 \mathrm{aX}$ :
$\mathrm{a}=\frac{1}{4}$
Focus =
$(X=a, Y=0)=\left(X=\frac{1}{4}, Y=0\right)=\left(x=\frac{1}{4}+1, y=1\right)=\left(x=\frac{5}{4}, y=1\right)$
Hence, the focus is at $\left(\frac{5}{4}, 1\right)$ 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. $\sqrt{5}$
- $\sqrt{5}$
Solution:
The equation of circle $x^2 + y^2 = 9$ The point is (2, c)
$\Rightarrow 2^2 + c^2= 9$
$4 + c^2 = 9$
$c^2 = 9 - 4$
$c^2 = 5$
$\text{c}=\sqrt{5}$ 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:
Answer
- 1
Solution:
Given, point $(1,2)$ and equation of circle is $x^2+y^2=5$
Now, $x^2+y^2-5=0$
Put $(1,2)$ in this equation, we get
$1^2+2^2-5=1+4-5=5-5=0$
So, the point $(1,2)$ lies on the circle.
only one tangent can be drawn. 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. $\Big(-\frac{1}{4},0\Big)$
- $\Big(-\frac{1}{4},0\Big)$
Solution:
Given:
Equation of the parabola = $y = 2x^2 + x$
$\Rightarrow\ \text{x}^2+\frac{\text{x}}{2}=\frac{\text{y}}{2}$
$\Rightarrow\ \Big(\text{x}+\frac{1}{4}\Big)^2=\frac{\text{y}}{2}+\frac{1}{16}$
$\Rightarrow\ \Big(\text{x}+\frac{1}{4}\Big)^2=\frac{8\text{y}+1}{16}$
$\Rightarrow\ \Big(\text{x}+\frac{1}{4}\Big)^2=\frac{1}{2}(\text{y}+\frac{1}{8})$
$\text{Let }\text{X}=\text{x}+\frac{1}{4},\text{Y}=\text{y}+\frac{1}{8}$
$\therefore\ \text{X}^2=\frac{1}{2}\text{Y}$
Comparing with X = 4aY
$\text{a}=\frac{1}{8}$
Focus $=(\text{X}=0,\ \text{Y}=\text{a})=\Big(\text{x}=\frac{-1}{4},\text{y}=0\Big)$
Hence, the focus is at $\Big(-\frac{1}{4},0\Big).$ 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. $x^2+y^2=16$
- $x^2+y^2=16$
Solution:
The equation of circle is $x^2+y^2=r^2$
Here, the radius is 4 So the equation is $x^2+y^2=4^2$
$x^2+y^2=16$ View full question & answer→MCQ 351 Mark
Equation of the hyperbola whose vertices are $( \pm 3,0)$ and foci at $( \pm 5,0)$, is
- ✓
$16 x^2-9 y^2=144$
- B
$9 x^2-16 y^2=144$
- C
$25 x^2-9 y^2=225$
- D
$9 x^2-25 y^2=81$
AnswerCorrect option: A. $16 x^2-9 y^2=144$
- $16 x^2-9 y^2=144$
Solution:
The vertices of the hyperbola are $( \pm 3,0)$ and foci are $( \pm 5,0)$.
Thus, the value of a and ae are 3 and 5 , respectively.
Now, using the relation $b^2=a^2\left(e^2-1\right)$, we get:
$b^2=25-9$
$\Rightarrow b^2=16$
Equation of hyperbola is given below:
$\frac{x^2}{9}-\frac{y^2}{16}=1$
$16 x^2-9 y^2=144$ View full question & answer→MCQ 361 Mark
The vertex of the parabola $x^2+8 x+12 y+4=0$ is
Answer
- (-4, 1)
Solution:
Given:
$x^2+8 x+12 y+4=0$
$\Rightarrow(x+4)^2-16+12 y+4=0$
$\Rightarrow(x+4)^2+12 y-12=0$
$\Rightarrow(x+4)^2=-12(y-1)$
Let $\mathrm{X}=\mathrm{x}+4, \mathrm{Y}=\mathrm{y}-1$
$X^2=-12 Y$
Vertex $=(X=0, Y=0)=(x+4=0, y-1=0)=(x=-4, y=1)$
Hence, the vertex is at $(-4,1)$. 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. $\text{g}^{2} + \text{f}^{2} – \text{c} \underline{>} 0$
- $\text{g}^{2} + \text{f}^{2} – \text{c} \underline{>} 0$
View full question & answer→MCQ 381 Mark
For what value of $k$, does the equation $9 x^2+y^2=k\left(x^2-y^2-2 x\right)$ represents equation of a circle?
Answer
- 4
Solution:
$9 x^2-k x^2+y^2+k y^2+2 k x=0$
$x^2(9-k)+y^2(1+k)+2 k x=0$
for circle
$9-k=1+k$
So, $k=4$ 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. $(x-6)^2+(y-5)^2=4$
- $(x-6)^2+(y-5)^2=4$
Solution:
If $(h, k)$ is the center and the radius is $r$ then the equation of the circle is given by
$(x-h)^2+(y-k)^2=r^2$
Given that The center of the circle $(h, k)=(6,5)$ and the radius $\mathrm{r}=2$
$\therefore$ The equation of the circle is $(x-6)^2+(y-5)^2=4$ 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. $\frac{4}{3}$
- $\frac{4}{3}$
Solution:
Since, the parabola $y^2 = 4ax$ passes through the point $(3, 2)$
$\Rightarrow 2^2 = 4a \times 3$
$\Rightarrow 4 = 12a$
$\Rightarrow\text{a} =\frac{ 4}{12}$
$\Rightarrow\text{a}=\frac{1}{3}$
So, the length of latusrectum $= \text{4a} = 4 \times (\frac{1}{3}) = \frac{4}{3}$ 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:
Answer
- 15
Solution:
The centre of the circle $x^2 + y^2 = 9$ is $(0, 0)$.
Let us denote it by $C_1$.
The centre of the circle $x^2 + y^2+ 8y + c = 0$ is $(0, -4)$.
Let us denote it by $C_2$.
The radius of $x^2 + y^2 = 9$ is $3$ units.
$x^2 + y^2+ 8y + c = 0$
$\Rightarrow(\text{x}-0)^2+(\text{y}+4)^2=16-\text{c}=(\sqrt{16-\text{c}})^2$
Therefore, the radius of the above circle is $\sqrt{16-\text{c}}$
Let the circles touch each other at P.
$\therefore\text{C}_1\text{C}_2=\text{PC}_2+\text{PC}_1$
$\Rightarrow\text{PC}_2=4-3=1$
$\Rightarrow\text{PC}_2-1=\sqrt{16-\text{c}}$
$\Rightarrow\text{c}=15$ 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
$x^2+y^2+a x+b y=0$
- ✓
$x^2+y^2-a x-b y=0$
- C
$x^2+y^2+b x+a y=0$
- D
AnswerCorrect option: B. $x^2+y^2-a x-b y=0$
- $x^2+y^2-a x-b y=0$
Solution:
Centre of the circle is $\Big(\frac{\text{a}}{2},\ \frac{\text{b}}{2}\Big)$ and its radius is $\sqrt{\Big(\frac{\text{a}}{2}\Big)^2+\Big(\frac{\text{b}}{2}\Big)^2}=\frac{1}{2}\sqrt{\text{a}^2+\text{b}^2}$
Equation of circle:
$\Big(\text{x}-\frac{\text{a}}{2}\Big)^2+\Big(\text{y}-\frac{\text{b}}{2}\Big)^2=\frac{1}{4}(\text{a}^2+\text{b}^2)$
$\Rightarrow(2\text{x}-\text{a}^2)+(2\text{y}-\text{b})^2=(\text{a}^2+\text{b}^2)$
$\Rightarrow4\text{x}^2+\text{a}^2-4\text{ax}+4\text{y}^2+\text{b}^2-4\text{by}=\text{a}^2+\text{b}^2$
$\Rightarrow\text{x}^2-\text{ax}+\text{y}^2-\text{by}=0$ 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. $2\text{e}_1^2 +\text{e}_2^2 = 3$
- $2\text{e}_1^2 +\text{e}_2^2 = 3$
Solution:
The standard from of the ellipse is $\frac{\text{x}^2}{18}+\frac{\text{y}^2}{4}=1,$ where $a^2 = 18$ and $b^2 = 4$.
So, the eccentricity is calculated in the following way:
$\text{b}2 = \text{a}2 (1 - \text{e}_1^2)$
$\Rightarrow4 = 18 (1 - \text{e}_1^2)$
$\Rightarrow\frac{2}{9}=1-\text{e}_1^2$
$\Rightarrow\text{e}_1^2=\frac{7}{9}$
The standard from of the hyperbola is $\frac{\text{x}^2}{9}-\frac{\text{y}^2}{4}=1,$ where $a^2 = 9$ and $b^2 = 4$.
So, the eccentricity is calculated in the following way:
$\text{b}^2 = \text{a}^2(\text{e}_2^2 - 1)$
$\Rightarrow4 = 9(\text{e}_2^2 - 1)$
$\Rightarrow\frac{4}{9}=\text{e}_2^2-1$
$\Rightarrow\text{e}_2^2=\frac{13}{9}$
$\therefore2\text{e}_1^2+\text{e}_2^2=\frac{2\times7}{9}+\frac{13}{9}$
$=\frac{27}{9}$
$=3$ 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:
Answer
- (-5, 10)
Solution:
Given the equation of the circle is $x^2+y^2+10 x-20 y+100=0$
or, $x^2+10 x+25+y^2-20 y+100=25$
or, $(x+5)^2+(y-10)^2=52$
From this equation it is clear that the centre is $(-5,10)$ View full question & answer→MCQ 461 Mark
The vertex of the parabola $(y-2)^2=16(x-1)$ is
Answer
- (1, 2)
Solution:
Given:
$(y-2)^2=16(x-1)$
$\text { Let } X=x-1, Y=y-2$
$\therefore Y^2=16 X$
$\text { Vertex }=(X=0, Y=0)=(x-1=0, y-2=0)=(x=1, y=2)$
Hence, the vertex is at $(1,2)$. 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. $4\sqrt2\text{a}$
- $4\sqrt2\text{a}$
Solution:
Let $O P$ be the chord.
Let the coordinates of $P$ be $\left(\mathrm{x}_1, \mathrm{y}_1\right)$.
From the figure, we have:
$O P^2=x_1^2+y_1^2 \ldots(1)$
And, $\tan \frac{\pi}{4}=\frac{y_1}{x_1}$
$\Rightarrow x_1=y_1 \ldots(2)$
Also, $\left(x_1, y_1\right)$ lies on the parabola.
$\therefore y_1{ }^2=4 \mathrm{ax}_{1 \ldots(3)}$
Using (2) and (3):
$x_1^2=4 a x_1 \Rightarrow x_1=4 a$
From (4), (1) and (2), we have:
$\mathrm{OP}^2=(4 a)^2+(4 a)^2=32 a^2$
$\Rightarrow \mathrm{OP}=4 \sqrt{2} \mathrm{a}$
Therefore, the length of the chord is $4 \sqrt{2} a$ a units. 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
AnswerGiven:
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:
$(\text{y}+1)=\frac{-3}{2}(\text{x}+4)$
$\Rightarrow\ 3\text{x}+2\text{y}+14=0$
View full question & answer→