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:
AnswerCorrect option: A. $7 x^2+2 x y+7 y^2-10 x+10 y+7=0$
- $7 x^2+2 x y+7 y^2-10 x+10 y+7=0$
Solution:
Given that, foums of the ellipse is $S(1,-1)$ and the equation of directrix is $x-y-3=0$
Also, $e=\frac{1}{2}$
From definition of ellipse, for any point $P(x, y)$ on the ellipse, we have $S P=e P M$, where $M$ is foot of the perpendicular from point $P$ to the directrix.
$\therefore \sqrt{(x-1)^2+(y+1)^2}=\frac{1}{2} \frac{|x-y-3|}{\sqrt{2}}$
$\Rightarrow 8 x^2-16 x+16+8 y^2+16 y=x^2+y^2+9-2 x y+6 y-6 x$
$\Rightarrow 7 x^2+2 x y+7 y^2-10 x+10 y+7=0$ View full question & answer→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
$\frac{2}{\sqrt5}$
AnswerCorrect option: B. ${\frac{\sqrt5}{2}}$
- ${\frac{\sqrt5}{2}}$
Solution:
The equation of the hyperbola is $x^2 - 4y^2 = 1$.
This can be rewritten in the following way:
$\frac{\text{x}^2}{1}-\frac{\text{y}^2}{\frac{1}{4}}=1$
This is the standard form of a hyperbola, where a = 1 and $\text{b}^2=\frac{1}{4}.$
The value of eccentricity is calculated in the following way:
$\text{b}^2=\text{a}^2(\text{e}^2-1)$
$\Rightarrow\frac{1}{4}=(\text{e}^2-1)$
$\Rightarrow\text{e}^2=\frac{5}{4}$
$\Rightarrow\text{e}=\frac{\sqrt5}{4}$ View full question & answer→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:
Answer$9\text{x}^2+16\text{y}^2=144$
$\Rightarrow\frac{\text{x}^2}{16}+\frac{\text{y}^2}{9}=1$
Equation of the tangent in case of an ellipse is given by
$\text{y}=\text{mx}+\sqrt{\text{a}^2\text{m}^2+\text{b}^2}$
$\Rightarrow\text{y}=\text{mx}+\sqrt{16\text{m}^2+9}\ \dots(1)$
Substituting x = 2 and y = 3, we get:
$3=2\text{m}\pm\sqrt{16\text{m}^2+9}$
$\Rightarrow3-2\text{m}=\sqrt{16\text{m}^2+9}$
On squaring both sides, we get:
$(3-2\text{m})^2=(16\text{m}^2+9)$
$\Rightarrow9+4\text{m}^2-12\text{m}=(16\text{m}^2+9)$
$\Rightarrow12\text{m}^2+12\text{m}=0$
$\Rightarrow12\text{m}(\text{m+1})=0$
$\Rightarrow\text{m}=0,-1$
Substituting values of m in eq. (1), we get:
For $\text{m}=0,\ \text{y}=3$
For $\text{m}=-1,\ \text{y}=-\text{x}+5$ or $\text{x}+\text{y}=5$
View full question & answer→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
$x^2+y^2+2 y=0$
AnswerCorrect option: C. $x^2+y^2-8 y+15=0$
- $x^2+y^2-8 y+15=0$
Solution:
Foot of perpendicular of $(5,4)$ on $y$-axis is $(0,4)$
$\therefore$ The equation of circle with
radius 1 cm is $(x-0)^2+(y-4)^2=1$
$\Rightarrow x^2+y^2-8 y+16$
$\Rightarrow x^2+y^2-8 y+15=0$ View full question & answer→MCQ 551 Mark
Choose the correct answer. The area of the circle centred at (1, 2) and passing through (4, 6) is:
AnswerCorrect 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 $=\pi\text{r}^2$
$=\pi\times(5)^2=25\pi$
View full question & answer→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
$17\pi$
AnswerCorrect option: B. $225\pi$
View full question & answer→MCQ 571 Mark
Equation of the diameter of the circle $x^2 + y^2 − 2x + 4y = 0$ which passes through the origin is:
Answer
- 2x + y = 0
Solution:
Let the diameter of the circle be y = mx.
Since the diameter of the circle passes through its centre, (1, -2) satisfies the equation of the diameter.
$\therefore$ m = -2
Substituting the value of m in the equation of diameter:
y = -2x
⇒ 2x + y = 0
Hence, the required equation of the diameter is 2x + y = 0. View full question & answer→MCQ 581 Mark
The length of the transverse axis is the distance between the:
AnswerThe length of the transverse axis is the distance between two vertices.
View full question & answer→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:
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
$2\sqrt{2}$
- ✓
AnswerThe 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 $=\sqrt{(-1)^2+\Big(\frac{3}{2}\Big)^2+\frac{5}{2}}=\sqrt{1+\frac{9}{4}+\frac{5}{2}}=\sqrt{\frac{23}{4}}=\frac{\sqrt{23}}{2}$
View full question & answer→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
$\pm1$
AnswerCorrect option: B. $\pm4$
- $\pm4$
Solution:
The equation of the circle is $x^2 + y^2 + 2ax + 8y + 16 = 0$.
Its centre is (-a, -4) and its radius is a units.
Since the circle touches the x-axis, we have:
$\sqrt{(-\text{a}+\text{a})^2+(4-0)^2}=\text{a}$
$\Rightarrow\text{a}=\pm4$ View full question & answer→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
AnswerCorrect 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 $\sqrt{\Big(\frac{3}{2}\Big)^2+(-1)^2-1}=\frac{3}{2}.$
View full question & answer→MCQ 631 Mark
The equation of a hyperbola with foci on the $x-$axis is:
- A
$\frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2} = 1$
- ✓
$\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2} = 1$
- C
$\text{x}^2 + \text{y}^2 = (\text{a}^2 + \text{b}^2)$
- D
$\text{x}^2 - \text{y}^2 = (\text{a}^2 + \text{b}^2)$
AnswerCorrect option: B. $\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2} = 1$
$\frac{\text{x}^2}{\text{a}^2}-\frac{\text{y}^2}{\text{b}^2} = 1$
View full question & answer→MCQ 641 Mark
The equation $x^2+ y^2 - 2x + 4y + 5 = 0$ represents:
- ✓
- B
- C
A circle of non zero radius
- D
Answer
- A point
Solution:
$x^2+y^2-2 x+4 y+5=0$
$(x-1)^2+(y+2)^2-5+5=0$
$\Rightarrow(x-1)^2+(y+2)^2=0$
Since, radius is 0, its a point
Alternative method:
Here, a = b = 1
$\text{r}=\sqrt{1+4-5=0}$
a circle of radius 0. So, its a point. View full question & answer→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
$x^2+y^2-4 x-2 y-4=0$
AnswerCorrect option: C. $x^2+y^2-4 x-2 y+4=0$
- $x^2+y^2-4 x-2 y+4=0$
Solution:
distance of pt. (2, 1) from line 3x + 4y - 5 is radius(r)
$\Rightarrow\text{r}=\frac{\mid6+4-5\mid}{5}=\frac{5}{5}=1$
⇒ Equation of circle is
⇒ $(x - 2)^2 + (y - 1)^2 = 1$
⇒ $x^2 + y^2 - 4x - 2y + 4 = 0$ View full question & answer→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
$6\sqrt2$
AnswerCorrect 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 $=\frac{2\times64}{\sqrt{64+64}}$
$=\frac{128}{8\sqrt2}$
$=\frac{16}{\sqrt2}$
$=8\sqrt2$
View full question & answer→MCQ 671 Mark
The equation of the conic with focus at (1, -1) directrix along x - y + 1 = 0 and eccentricity $\sqrt2$ is
Answer
- 2xy - 4x + 4y + 1 = 0
Solution:
Let P(x, y) be any point on the hyperbola.
Then, the distance of any point from the focus is eccentricity times the distance from the directrix.
$\therefore\sqrt{(\text{x}-1)^2+(\text{y}+1)^2}=\sqrt2\Big|\frac{\text{x}-\text{y}+1}{\sqrt2}\Big|$
Squaring both the sides, we get:
$(x - 1)^2 + (y + 1)^2 = (x - y + 1)^2$
$x^2 - 2x + 1 + y^2 + 1 + 2y = x^2 + y^2 + 1 - 2xy - 2y + 2x$
$2xy - 4x + 4y + 1 = 0$ View full question & answer→MCQ 681 Mark
The equation of a circle with radius $5$ and touching both the coordinate axes is:
- A
$x^2+y^2 \pm 10 x \pm 10 y+5=0$
- B
$x^2+y^2 \pm 10 x \pm 10 y=0$
- ✓
$x^2+y^2 \pm 10 x \pm 10 y+25=0$
- D
$x^2+y^2 \pm 10 x \pm 10 y+51=0$
AnswerCorrect option: C. $x^2+y^2 \pm 10 x \pm 10 y+25=0$
- $x^2+y^2 \pm 10 x \pm 10 y+25=0$
Solution:
Case I: If the circle lies in the first quadrant:
The equation of a circle that touches both the coordinate axes and hasradius a is $x^2+y^2-2 a x-2 a y+a^2=0$.
The given radius of the circle is 5 units, i.e. $a=5$.
Thus, the equation of the circle is $x^2+y^2-10 x-10 y+25=0$.
Case II: If the circle lies in the second quadrant:
The equation of a circle that touches both the coordinate axes and has radius $a$ is $x^2+y^2+2 a x-2 a y+a^2=0$.
The given radius of the circle is 5 units, i.e. $a=5$.
Thus, the equation of the circle is $x^2+y^2+10 x-10 y+25=0$.
Case III: If the circle lies in the third quadrant:
The equation of a circle that touches both the coordinate axes and has radius $a$ is $x^2+y^2+2 a x+2 a y+a^2=0$
The given radius of the circle is 5 units, i.e. $a=5$.
Thus, the equation of the circle is $x^2+y^2+10 x+10 y+25=0$.
Case IV: If the circle lies in the fourth quadrant:
The equation of a circle that touches both the coordinate axes and has radius $a$ is $x^2+y^2-2 a x+2 a y+a^2=0$.
The given radius of the circle is 5 units, i.e. $a=5$.
Thus, the equation of the circle is $x^2+y^2-10 x+10 y+25=0$.
Hence, the required equation of the circle is $x^2+y^2 \pm 10 x \pm 10 y+25=0$. View full question & answer→MCQ 691 Mark
The perpendicular distance from the point (3, -4) to the line 3x - 4y + 10 = 0:
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:
AnswerCorrect option: A. $(-1,\ 3)$
- $(-1,\ 3)$
Solution:
The given equation of the curve is $x^2 + y^2 = 25$
Since $(\lambda,\ \lambda+1)$ lies inside the region bounded by the curve $x^2 + y^2 = 25$ and the y-axis, we have:
$\lambda^2+(\lambda+1)^2<25,$ provided $\lambda+1>0$
$\Rightarrow\lambda^2+\lambda^2+12\lambda<25,\ \lambda>-1$
$\Rightarrow2\lambda^2+2\lambda-24<0,\ \lambda>-1$
$\Rightarrow\lambda^2+\lambda-12<0,\ \lambda>-1$
$\Rightarrow(\lambda-3)(\lambda+4)<0,\ \lambda>-1$
$\Rightarrow-4<\lambda<3,\ \lambda>-1$
$\Rightarrow\lambda\in(-1,\ 3)$ View full question & answer→MCQ 711 Mark
The center of the circle $4 x^2+4 y^2-8 x+12 y-25=0$ is:
View full question & answer→MCQ 721 Mark
The equation of circle center at (0, 0) and Radius 8cm:
- ✓
$x^2+y^2=64 cm$
- B
$x^2+y^2=8$
- C
$x^2+y^2=16$
- D
$x^2+y^2=4$
AnswerCorrect option: A. $x^2+y^2=64 cm$
- $x^2+y^2=64 cm$
Solution:
The equation of circle is $x^2+y^2=r^2$
$x^2+y^2=8^2$
$x^2+y^2=64$ View full question & answer→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
AnswerCorrect option: A. $\frac{225\sqrt{3}}{6}$
- $\frac{225\sqrt{3}}{6}$
Solution:
Let ABC be the required equilateral triangle.
The equation of the circle is $x^2 + y^2 - 6x - 8y - 25 = 0$.
Therefore, coordinates of the centre O is (3, 4).
Radius of the circle $=\text{OA}=\text{OB}=\text{OC}=\sqrt{9+16+25}=5\sqrt{2}$
In $\Delta\text{BOD},$ we have:
$\sin60^\circ=\frac{\text{DB}}{\text{BO}}$
$\Rightarrow\text{DB}=\frac{\sqrt{3}}{2}(5\sqrt{2})$
$\Rightarrow\text{BC}=2\text{BD}-\sqrt{3}\big(5\sqrt{2}\big)=5\sqrt{6}$
Now, area of $\triangle\text{ABC}=\frac{\sqrt{3}}{4}\text{BC}^2=\big(5\sqrt{6}\big)^2\\=\frac{\sqrt{3}(150)}{4}=\frac{\sqrt{3}(75)}{2}=\frac{\sqrt{3}(225)}{6}$ square units View full question & answer→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
$x^2+y^2=16$
- ✓
$x^2-y^2=32$
AnswerCorrect option: D. $x^2-y^2=32$
- $x^2-y^2=32$
Solution:
The distance between the foci is 2ae.
$\therefore$ 2ae = 16
⇒ ae = 8
$\text{e}=\sqrt2$
$\therefore\text{a}\sqrt2=8$
$\Rightarrow\text{a}=4\sqrt2$
Also, $b^2 = a^2(e^2 − 1)$
$\Rightarrow b^2 = 32(2 − 1)$
$\Rightarrow b^2 = 32$
Standard form of the hyperbola is given below:
$\frac{\text{x}^2}{32}-\frac{\text{y}^2}{32}=1$
$\text{x}^2-\text{y}^2=32$ View full question & answer→MCQ 751 Mark
Which of the following points lie on the parabola $x^2=4 a y$ ?
- A
$x=a t^2, y=2 a t$
- B
$x=2 a t, y=a t^2$
- C
$x=2 a t^2, y=a t$
- ✓
$x=2 a t, y=a t^2$
AnswerCorrect option: D. $x=2 a t, y=a t^2$
- $x=2 a t, y=a t^2$
Solution:
Substituting $x=2 a t, y=a t^2$ in the given equation:
$(2 a t)^2=4 a\left(a t^2\right)$
$\Rightarrow 4 a^2 t^2=4 a^2 t^2$
Hence, (2at, at ${ }^2$ ) lies on the parabola $x^2=4 a y$. View full question & answer→MCQ 761 Mark
The equation of the circle concentric with $x^2+y^2-3 x+4 y-c=0$ and passing through $(-1,-2)$ is:
- A
$x^2+y^2-3 x+4 y-1=0$
- ✓
$x^2+y^2-3 x+4 y=0$
- C
$x^2+y^2-3 x+4 y+2=0$
- D
AnswerCorrect option: B. $x^2+y^2-3 x+4 y=0$
- $x^2+y^2-3 x+4 y=0$
Solution:
The centre of the circle $x^2 + y^2 - 3x + 4y - c = 0$ is $\Big(\frac{3}{2},\ -2\Big).$
Therefore, the centre of the required circle is $\Big(\frac{3}{2},\ -2\Big).$
The equation of the circle is $\Big(\text{x}-\frac{3}{2}\Big)^2+(\text{y}+2)^2=\text{a}^2. \ ......(1)$
Also, circle (1) passes through (-1, -2).
$\therefore\Big(-1-\frac{3}{2}\Big)^2+\Big(-2+2\Big)^2=\text{a}^2$
$\Rightarrow\text{a}=\frac{5}{2}$
Substituting the value of a in equation (1):
$\Big(\text{x}-\frac{3}{2}\Big)^2+(\text{y}+2)^2=\Big(\frac{5}{2}\Big)^2$
$\Rightarrow\frac{(2\text{x}-3)^2}{4}+(\text{y}+2)^2=\frac{25}{4}$
$\Rightarrow(2\text{x}-3)^2+4(\text{y}+2)^2=25$
$\Rightarrow\text{x}^2+\text{y}^2-3\text{x}+4\text{y}=0$
Hence, the required equation of the circle is $x^2 + y^2 - 3x + 4y = 0$. View full question & answer→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
$x^2+y^2=16$
AnswerCorrect option: B. $x^2+y^2-2 x-4 y=0$
- $x^2+y^2-2 x-4 y=0$
Solution:
Given, points are (2, 0) and (0, 4)
$\therefore$ equation of circle is (x - 2) (x - 0) + (y - 0) (y - 4) = 0
By expanding, we get
$x^2 - 2x + y^2 - 4y = 0$ View full question & answer→MCQ 781 Mark
Choose the correct answer. The equation of a circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length 3a is:
[Hint: Centroid of the triangle coincides with the centre of the circle and the radius of the circle is $\frac{2}{3}$ of the length of the mediam]
- A
$x^2+y^2=9 a^2$
- B
$x^2+y^2=16 a^2$
- ✓
$x^2+y^2=4 a^2$
- D
$x^2+y^2=a^2$
AnswerCorrect option: C. $x^2+y^2=4 a^2$
- $x^2+y^2=4 a^2$
Solution:
Let ABC be an equilateral triangle in which mediam AD = 3a
Centre of the circle is same as the centroid of the triangle i.e., $(0, 0)$
AG : GD = 2 : 1
So, $\text{AG}=\frac{2}{3}\text{AD}=\frac{2}{3}\times3\text{a}=2\text{a}$
$\therefore$ The equation of the circle is,
$(x - 0)^2 + (y - 0)^2= (2a)^2$
$\Rightarrow x^2 + y^2 = 4a^2$ View full question & answer→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
$x^2+y^2-4 x+2 y+45=0$
AnswerCorrect option: A. $x^2+y^2-4 x+2 y=45$
- $x^2+y^2-4 x+2 y=45$
Solution:
Equation of circle, $(\text{x} - 2)^2 + (\text{y} -( -1))^2= \Big(\sqrt{{(3-2)^2+(6}-(-1))^2\Big)}^2$
$\text{x}^2 - 4\text{x} + 4 + \text{y}^2 + 2\text{y} + 1=(\sqrt{1+49})^2\therefore\text{x}^2+\text{y}^2-4\text{x}+2\text{y}=45$ Equation of circle. View full question & answer→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
$\text{none of these}$
AnswerCorrect 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, $\text{e}\sqrt{1-\frac{\text{b}^2}{\text{a}^2}}$
$\Rightarrow\text{e}=\sqrt{1-\frac{\text{b}^2}{4\text{b}^2}}$
$\Rightarrow\text{e}=\sqrt{1-\frac{1}{4}}$
$\Rightarrow\text{e}=\frac{\sqrt{3}}{2}$
View full question & answer→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
AnswerCorrect option: D. Assertion is incorrect but Reason is correct
- Assertion is incorrect but Reason is correct
Solution:
$(x + 1)^2 + (y - 1)^2= 2^2$ Radius = 2 Centre (-1, 1) Assertion is incorrect but reason is correct. View full question & answer→MCQ 821 Mark
The equation of the circle which touches $x$-axis at $(0,0)$ and touches the line $3 x+4 y-5=0$ is:
- A
$x^2+y^2-4 y=0$
- B
$x^2+y^2-10 y=0$
- C
$x^2+y^2+10 x=0$
- ✓
$x^2+y^2+10 y=0$
AnswerCorrect option: D. $x^2+y^2+10 y=0$
- $x^2+y^2+10 y=0$
Solution:
Equation of circle touching $x$-axis at $(0,0)$, means centre of circle lie on $Y$-axis i.e. $(0, k)$.
$(x-0)^2+(y-k)^2=k^2$
$S: x^2+y^2-2 k y=0$.... (1)
Circle S touches $3 x+4 y-5=0$
$\therefore \mathbf{k}=\frac{4 \mathbf{k}-5}{5}$
$5 k=4 k-5$
$k=-5$
$\therefore$ Equation of circle is
$\Rightarrow x^2+y^2+10 y=0$ View full question & answer→MCQ 831 Mark
Choose the correct answer. Equation of a circle which passes through $(3,6)$ and touches the axes is:
- A
$x^2+y^2+6 x+6 y+3=0$
- B
$x^2+y^2-6 x-6 y-9=0$
- ✓
$x^2+y^2-6 x-6 y+9=0$
- D
AnswerCorrect option: C. $x^2+y^2-6 x-6 y+9=0$
- $x^2+y^2-6 x-6 y+9=0$
Solution:
Given that the circle touches both axes.
Therefore, equation of the circle is, $(x - a)^2 + (y - a)^2 = a^2$
Circle passes through the point (3, 6)
$\therefore (3 - a)^2 + (6 - a)^2 = a^2$
$\Rightarrow a^2 - 18a + 45 = 0$
$\Rightarrow (a - 3)(a - 15) = 0$
$\therefore a = 3, a = 15$
For a = 3, the equation of circle is,
$(x - 3)^2 + (y - 3)^2 = 9$
$\Rightarrow x^2 + y^2 - 6x - 6y + 9 = 0$ View full question & answer→MCQ 841 Mark
If $(-3,2)$ lies on the circle $x^2+y^2+2 g x+2 f y+c=0$ which is concentric with the circle $x^2+y^2+6 x+8 y-5=0$, then $\mathrm{c}=$
Answer
- -11
Solution:
The centre of the circle $x^2+y^2+6 x+8 y-5=0$ is $(-3,-4)$.
The circle $x^2+y^2+2 g x+2 f y+c=0$ is concentric with the circle $x^2+y^2+6 x+8 y-5=0$.
Thus, the centre of $x^2+y^2+2 g x+2 f y+c=0$ is $(-3,-4)$.
$\therefore g=3, f=4$
Also, it is given that $(-3,2)$ lies on the circle $x^2+y^2+2 g x+2 f y+c=0$.
$\therefore(-3)^2+2^2+2(3)(-3)+2(4)(2)+c=0$
$\Rightarrow 9+4-18+16+c=0$
$\Rightarrow c=-11$ View full question & answer→MCQ 851 Mark
Choose the correct answer. If the focus of a parabola is (0, -3) and its directrix is y = 3, then its equation is:
- ✓
$x^2=-12 y$
- B
$x^2=12 y$
- C
$y^2=-12 x$
- D
$y^2=12 x$
AnswerCorrect option: A. $x^2=-12 y$
- $x^2=-12 y$
Solution:
According to the definition of parabola,
$\sqrt{(\text{x}-0)^2+(\text{y}+3)^2}=\Bigg|\frac{\text{y}-3}{\sqrt{(0)^2+(1)^2}}\Bigg|$
$\Rightarrow\sqrt{\text{x}^2+\text{y}^2+9+6\text{y}}=|\text{y}-3|$
Squaring both sides, we get
$x^2+y^2+9+6 y=y^2+9-6 y$
$\Rightarrow x^2+9+6 y=9-6 y$
$\Rightarrow x^2=-12 y$ View full question & answer→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
$y^2=8(x+5)$
AnswerCorrect option: A. $y^2=8(x+3)$
- $y^2=8(x+3)$
Solution:
Given that vertex $\equiv(-3,0)$ and directrix, x + 5 = 0
So, focus $\equiv\text{S}(-1,0)$
For any point of parabola P(x, y) we have,
$\text{SP}=\text{PM}$
$\Rightarrow\sqrt{(\text{x}+1)+\text{y}^2}=|\text{x}+5|$
$\Rightarrow\text{x}^2+2\text{x}+1+\text{y}^2=\text{x}^2+10\text{x}+25$
$\Rightarrow\text{y}^2=8\text{x}+24$
$\Rightarrow\text{y}^2=8(\text{x}+3)$ View full question & answer→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
$(x+3)^2+(y-1)^2=7$
- ✓
$(x+3)^2+(y-1)^2=49$
AnswerCorrect option: D. $(x+3)^2+(y-1)^2=49$
- $(x+3)^2+(y-1)^2=49$
Solution:
The general equation of a circle with center at $(a, b)$ and radius $r$ is $(x-a)^2+(y-b)^2=r^2$
So substituting the values we get the equation of the circle is $(x+3)^2+(y-1)^2=7^2=49$ View full question & answer→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
AnswerCorrect option: C. $\sqrt{\frac{3}{2}}$
The lengths of the latus rectum and the transverse axis are $\frac{2\text{b}^2}{\text{a}}\text{ and }2\text{a},$ respectively.
According to the given statement, length of the latus rectum is half of its transverse axis.
$\therefore\frac{2\text{b}^2}{\text{a}}=\frac{1}{2}\times2\text{a}$
$\Rightarrow\frac{2\text{b}^2}{\text{a}}=\text{a}$
$\Rightarrow2\text{b}^2=\text{a}$
Eccentricity, $\text{e}=\frac{\sqrt{\text{a}^2+\text{b}^2}}{\text{a}}$
Substituting the value $\text{b}^2=\frac{\text{a}^2}{2},$ we get:
$\text{e}=\frac{\sqrt{\text{a}^+\frac{\text{a}}{2}}}{\text{a}}$
$=\frac{\text{a}\sqrt{\frac{3}{2}}}{\text{a}}$
$=\sqrt{\frac{3}{2}}$
$\therefore$ Eccentricity is $\sqrt{\frac{3}{2}}$
View full question & answer→MCQ 891 Mark
If $e_1$ is the eccentricity of the conic $9 x^2+4 y^2=36$ and $e_2$ is the eccentricity of the conic $9 x^2-4 y^2=36$, then
- A
$\text{e}_1^2-\text{e}_2^2=2$
- ✓
$2<\text{e}_2^2-\text{e}_1^2<3$
- C
$\text{e}_2^2-\text{e}_1^2=2$
- D
$\text{e}_2^2-\text{e}_1^2>3$
AnswerCorrect option: B. $2<\text{e}_2^2-\text{e}_1^2<3$
- $2<\text{e}_2^2-\text{e}_1^2<3$
Solution:
The conic $9x^2 + 4y^2 = 36$ can rewritten in the following way:
$\frac{9\text{x}^2}{36}+\frac{4\text{y}^2}{36}=1$
$\Rightarrow\frac{\text{x}^2}{4}+\frac{\text{y}^2}{9}=1$
This is the standard equation of an ellipse.
$\therefore b^2 = a^2(1−e_1)^2$
$\Rightarrow9=4(1-\text{e}_1)^2$
$\Rightarrow(\text{e}_1)^2=\frac{-5}{4}$
The conic $9x^2 − 4y^2 = 36$ can rewritten in the following way:
$\frac{9\text{x}^2}{36}-\frac{4\text{y}^2}{36}=1$
$\Rightarrow\frac{\text{x}^2}{4}-\frac{\text{y}^2}{9}=1$
This is the standard equation of a hyperbola.
$\therefore b^2 = a^2(e_2^2 − 1)$
$\Rightarrow9=4(\text{e}_2^2-1)$
$\Rightarrow(\text{e}_2)^2=\frac{13}{4}$
$\therefore\text{e}_2^2-\text{e}_1^2=\frac{13}{4}+\frac{5}{4}=2.5$ View full question & answer→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
$\frac{1}{\text{a}}+\frac{1}{\text{b}}=\frac{2}{\text{c}}$
AnswerCorrect option: A. $\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}=\frac{1}{\text{c}}$
- $\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}=\frac{1}{\text{c}}$
Solution:
Given:
$x^2 + y^2 + 2ax + c = 0$ ....... (1)
And, $x^2 + y^2 + 2by + c = 0$ ........ (2)
For circle (1), we have:
Centre = $(-a, 0) = C_1$
For circle (2), we have:
Centre = $(0,-b) = C_2$
Let the circles intersect at point P.
$\therefore$ Coordinates of P = Mid point of $C_1C_2$
$\Rightarrow$ Coordinates of P $=\Big(\frac{-\text{a}+0}{2},\ \frac{0-\text{b}}{2}\Big)=\Big(\frac{-\text{a}}{2},\ \frac{-\text{b}}{2}\Big)$
Now, we have:
$PC_1$ = radius of (1)
$\Rightarrow\sqrt{(-\text{a}+\frac{\text{a}}{2})^2}+\Big(0-\frac{\text{b}}{2}\Big)^2=\sqrt{\text{a}^2-\text{c}}$
$\Rightarrow\frac{\text{a}^2}{4}+\frac{\text{b}}{4}^2=\text{a}^2-\text{c}\ .....(3)$
Also, radius of circle (1) = radius of circle (2)
$\Rightarrow\sqrt{\text{a}^2-\text{c}}=\sqrt{\text{b}^2-\text{c}}$
$\Rightarrow\text{a}^2=\text{b}^2\ .....(4)$
From (3) and (4), we have:
$\frac{\text{a}^2}{2}=\text{a}^2-\text{c}$
$\Rightarrow\frac{\text{a}^2}{2}=\text{c}$
$\Rightarrow\frac{2}{\text{a}^2}=\frac{1}{\text{c}}$
$\Rightarrow\frac{1}{\text{a}^2}+\frac{1}{\text{a}^2}=\frac{1}{\text{c}}$
$\Rightarrow\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}=\frac{1}{\text{c}}$ View full question & answer→MCQ 911 Mark
For the ellipse $12\text{x}^2+4\text{y}^2+24\text{x}-16\text{y}+25=0$
AnswerDisclaimer : The equation should be $12\text{x}^2+4\text{y}^2+24\text{x}-16\text{y}+24=0$ instead of $12\text{x}^2+4\text{y}^2+24\text{x}-16\text{y}+25=0.$
$12\text{x}^2+4\text{y}^2+24\text{x}-16\text{y}+24=0$
$\Rightarrow12\big(\text{x}^2+2\text{x}\big)+4\big(\text{y}^2-4\text{y}\big)=-24$
$\Rightarrow12\big(\text{x}^2+2\text{x}+1\big)+4\big(\text{y}^2-4\text{y}+4\big)=-24+12+16$
$\Rightarrow12\big(\text{x}+1\big)^2+4\big(\text{y}-2\big)^2=4$
$\Rightarrow\frac{(\text{x}+1)^2}{3}+\frac{(\text{y}-2)^2}{1}=1$
So, the centre is a $(-1,\ 2).$
Here, $\text{a}=\sqrt{3}$ and $\text{b}=1$
The lengths of the axes are $\sqrt{3}$ and 1.
Now, $\text{e}=\sqrt{1-\frac{\text{b}62}{\text{a}^2}}$
$\text{e}=\sqrt{1-\frac{1}{3}}$
$\Rightarrow\text{e}=\sqrt{\frac{2}{3}}$
View full question & answer→MCQ 921 Mark
The parametric equations of a parabola are $x=t^2+1, y=2 t+1$. The cartesian equation of its directrix is
Answer
- x = 0
Solution:
Given:
$x=t^2+1$
$y=2 t+1$
From (1) and (2):
$x=\left(\frac{y-1}{2}\right)^2+1$
On simplifying:
$(y-1)^2=4(x-1)$
Let $\mathrm{Y}=\mathrm{y}-1$ and $\mathrm{X}=\mathrm{x}-1$
$\therefore Y^2=4 X$
Comparing it with $y^2=4 a x:$
$a=1$
Therefore, the equation of the directrix is $X=-a$, i.e. $x-1=-1 \Rightarrow x=0$ View full question & answer→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
$4$
AnswerCorrect option: B. $\frac{4}{3}$
- $\frac{4}{3}$
Solution:
Given parabola is $y^2 = 4ax$
If the parabola is passing through (3, 2)
Then $(2)^2 = 4a \times 3$
$\Rightarrow 4 = 12a$
$\Rightarrow\text{a}=\frac{1}{3}$
Nowm length of the latus rectum $=4\text{a}=4\times\frac{1}{3}=\frac{4}{3}$ View full question & answer→MCQ 941 Mark
The focus of the parabola $y^2 = 8x$ is:
Answer
- (2, 0)
Solution:
Given parabola equation $y^2 = 8x …(1)$
Here, the coefficient of x is positive and the standard form of parabola is $y^2 = 4ax … (2)$
Comparing (1) and (2), we get
4a = 8
$\text{a} = \frac{8}{4} = 2$
We know that the focus of parabolic equation $y^2 = 4ax$ is $(a, 0)$.
$\therefore$ The focus of the parabola $y^2 = 8x$ is $(2, 0)$. View full question & answer→MCQ 951 Mark
The equation of the directrix of the parabola whose vertex and focus are (1, 4) and (2, 6) respectively is
AnswerGiven:
The vertex and the focus of a parabola are (1, 4) and (2, 6), respectively.
$\therefore$ Slope of the axis of the parabola $= \frac{6-4}{2-1}=2$
Slope of the directrix $=\ \frac{-1}{2}$
Let the directrix intersect the axis at K (r, s).
$\therefore\ \frac{\text{r}+2}{2}=1,\ \frac{\text{s}+6}{2}=4$
$\Rightarrow\ \text{r}=0,\ \text{s}=2$
Equation of the directrix:
$(\text{y}-2)=\frac{-1}{2}(\text{x}-0)$
⇒ x + 2y = 4
View full question & answer→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
$(y+2)^2=16(x-1)$
AnswerCorrect option: C. $(y-2)^2=16(x+1)$
View full question & answer→MCQ 971 Mark
Latus rectum of a parabola is a $........$ line segment with respect to the axis of the parabola through the focus whose endpoints lie on the parabola:
AnswerConsider the above image which shows that the latus rectum of a parabola is a line segment perpendicular to the axis of the parabola, through the focus and whose end points lie on the parabola.
View full question & answer→MCQ 981 Mark
If the eccentricity of the hyperbola $x^2-y^2 \sec ^2 \alpha=5$ is $\sqrt{3}$ times the eccentricity of the ellipse $x^2 \sec ^2 \alpha+y^2=25$, then $\alpha=$
- A
$\frac{\pi}{6}$
- ✓
$\frac{\pi}{4}$
- C
$\frac{\pi}{3}$
- D
$\frac{\pi}{2}$
AnswerCorrect option: B. $\frac{\pi}{4}$
- $\frac{\pi}{4}$
Solution:
The hyperbola $\text{x}^2 − \text{y}^2 \sec^2\alpha = 5$ can be rewritten in the following way:
$\frac{\text{x}^2}{5}-\frac{\text{y}^2}{5\cos^2\text{a}}=1$
This is the standard form of a hyperbola, where $ \text{a}^2 = 5 \text{ and } \text{b}^2 = 5\cos^2\alpha.$
$\Rightarrow\text{b}^2 = \text{a}^2(\text{e}_1^2 − 1)$
$⇒ 5\cos^2\alpha=5(\text{e}_1^2−1)$
$\Rightarrow\text{e}_1^2=\cos^2\alpha+1...(1)$
The ellipse $\text{x}^2\sec^2\alpha+\text{y}^2=25$ can be rewritten in the following way:
$\frac{\text{x}^2}{25\cos^2\alpha}+\frac{\text{y}^2}{25}=1$
This is the standard form of an ellipse, where $\text{a}^2=25\text{ and }\text{b}^2=25\cos^2\alpha$
$\text{b}^2=\text{a}^2(1-\text{e}_2^2)$
$\Rightarrow\text{e}_2^2=1-\cos^2\alpha$
$\Rightarrow\text{e}_2^2=\sin^2\alpha...(2)$
According to the question,
$\cos^2\alpha+1=3(\sin^2\alpha)$
$\Rightarrow2=4\sin^2\alpha$
$\Rightarrow\sin\alpha=\frac{1}{\sqrt2}$
$\Rightarrow\alpha=\frac{\pi}{4}$ View full question & answer→MCQ 991 Mark
Find the equation to the circle which touches the axis of $y$ at the origin and passes through the point (b, c):
- A
$b x^2+b y^2-\left(b^2+c^2\right) y=0$
- ✓
$b x^2+b y^2-\left(b^2+c^2\right) x=0$
- C
$b x^2+b y^2+\left(b^2+c^2\right) y=-1$
- D
$b x^2+c y^2-\left(b^2+c^2\right) x=1$
AnswerCorrect option: B. $b x^2+b y^2-\left(b^2+c^2\right) x=0$
- $b x^2+b y^2-\left(b^2+c^2\right) x=0$
Solution:
Equation of circle which touches the $y$-axis at origin is $x^2+y^2+2 g x+d=0$
Since the circle passes through origin, $d=0$ Thus the equation becomes, $x^2+y^2+2 g x=0 \ldots .$. (1)
The equation passes through $(b, c)$
so, $b^2+c^2+2 g b=0$
$\therefore \mathrm{g}=\frac{-\mathrm{b}^2-\mathrm{c}^2}{2 \mathrm{~b}}$
So, putting the value of $g$ in (1) we getb $x^2+b y^2-\left(b^2+c^2\right) x=0$ View full question & answer→MCQ 1001 Mark
The equation of parabola with vertex at origin and directrix $x-2=0$ is:
- A
$y^2=-4 x$
- B
$y^2=4 x$
- ✓
$y^2=-8 x$
- D
$y^2=8 x$
AnswerCorrect option: C. $y^2=-8 x$
View full question & answer→