Free Conic Sections 02 Practice Test - 11th Grade - Commerce

Question 1

If a circle whose centre is (1, –3) touches the line 3x – 4y –5 = 0, then the radius of the circle is

$\frac{5}{2}$

$\frac{7}{2}$

SOLUTION

Solution : A

Radius = perpendicular distance from (1, -3) to the given line 3x - 4y - 5 = 0.
i.e. $∣ ∣ ∣ \frac{3 + 12 - 5}{\sqrt{5^{2}}} ∣ ∣ ∣ = 2.$

Question 2

Consider a family of circles which are passing through the point (–1, 1) and are tangent to x-axis. (h, k) are the co-ordinates of the centre of the circles, then the set of values of k is given by the interval

$- \frac{1}{2} \leq k \leq \frac{1}{2}$

$k \leq \frac{1}{2}$

< k < $\frac{1}{2}$

$k \geq \frac{1}{2}$

SOLUTION

Solution : D

Since, circles are passing through the point (–1, 1) and touching X-axis, the equationof circles can be written as -
$(h + 1)^{2} + (K - 1)^{2} = K^{2}$
$h^{2} + 2 h + 2 - 2 K = 0$
If h is real the $Δ \geq 0$
4- 4.1. (2 – 2K) $\geq 0$
= 4 - 8 + 8k $\geq 0$
= 8k $\geq 4$
= $k \geq \frac{1}{2}$

Question 3

The end points of latus rectum of the parabola $x^{2} = 4 a y$ are

(a,2a) , (2a, -a)

(-a,2a) , (2a, a)

(a, - 2a) , (2a, a)

(-2a , a) , (2a, a)

SOLUTION

Solution : D

It is a fundamental concept. The end points of latus rectum of the parabola $x^{2} = 4 a y$ are (-2a , a) , (2a, a).

Question 4

The ends of latus rectum of parabola $x^{2} + 8 y = 0$ are

(-4, -2) and (4, 2)

(4, -2) and (-4, 2)

(-4, -2) and (4, -2)

(4, 2) and (-4, 2)

SOLUTION

Solution : C

$x^{2} = - 8 y \Rightarrow a = - 2 .$ So , focus = (0,-2)

Ends of latus rectum = (4,-2) , (-4,-2) .

Trick: Since the ends of latus rectum lie on parabola , so only points (-4,-2) and (4,-2) satsify the parabola.

Question 5

The focus of the parabols $x^{2} = - 16 y$ is

(4, 0)

(0, 4)

(-4, 0)

(0, -4)

SOLUTION

Solution : D

a = 4 , vertex = (0,0) , focus = (0,-4) .

Question 6

The points on the parabola $y^{2} = 36 x$ whose ordinate is three times the absicca are

(0, 0), (4, 12)

(1, 3), (4, 12)

(4, 12)

None of these

SOLUTION

Solution : A

$y_{1} = 3 x_{1}$ . According to given condition $9 x_{1}^{2} = 36 x_{1}$

$\Rightarrow x_{1} = 4, 0 \Rightarrow y_{1} = 12, 0$

Hence the points are (0,0) and (4,12).

Question 7

If the semi-major axis of an ellipse is 3 and the latus rectum is 16/9, then the standard equation of the ellipse is

$\frac{x^{2}}{9} + \frac{y^{2}}{8} = 1$

$\frac{x^{2}}{9} + \frac{3 y^{2}}{8} = 1$

$\frac{x^{2}}{9} + \frac{8 y^{2}}{3} = 1$

$\frac{x^{2}}{3} + \frac{y^{2}}{8} = 1$

SOLUTION

Solution : B

Given that a = 3
$L a t u s R e c t u m = \frac{2 b^{2}}{a} = \frac{16}{9} \frac{2}{3} b^{2} = \frac{16}{9} b^{2} = \frac{8}{3}$

The standard equation of the ellipse is

$\frac{x^{2}}{9} + \frac{3 y^{2}}{8} = 1$

Question 8

Let S and T be the foci of the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{8} = 1$ . If P(x,y) is any point on the ellipse, then the maximum area of the triangle PST (in square units) is ___

$2 \sqrt{2}$

$4 \sqrt{2}$

$8 \sqrt{2}$

SOLUTION

Solution : C

The given ellipse is
$\frac{x^{2}}{16} + \frac{y^{2}}{8} = 1 a^{2} = 16; b^{2} = 8 Focal length, c = \sqrt{16 - 8} = \sqrt{8}$

The foci of the ellipse lie on the major axis. The greatest perpendicular distance from the major axis to any point on the ellipse is the length of the semi-minor axis.

For the triangle PST,
$S T = 2 S = 2 \sqrt{8} Max Area = \frac{1}{2} \times S T \times b = \frac{1}{2} \times 2 \sqrt{8} \times \sqrt{8} = 8 sq. units$

Question 9

The eccentricity of the ellipse $12 x^{2} + 7 y^{2} = 84$ is equal to

$\frac{\sqrt{5}}{7}$

$\sqrt{\frac{5}{12}}$

$\frac{\sqrt{5}}{12}$

$\sqrt{\frac{5}{7}}$

SOLUTION

Solution : B

$12 x^{2} + 7 y^{2} = 84 \frac{x^{2}}{7} + \frac{y^{2}}{12} = 1 a^{2} = 7; b^{2} = 12 F o c a l l e n g t h, c = \sqrt{b^{2} - a^{2}} c = \sqrt{12 - 7} = \sqrt{5} e = \frac{c}{a} = \sqrt{\frac{5}{12}}$

Question 10

For each point (x,y) on an ellipse, the sum of the distances from (x,y) to the points (2,0) and (-2,0) is 8. Then the positive value of x so that (x,3) lies on the ellipse is ___

$2 \sqrt{3}$

$\frac{1}{\sqrt{3}}$

SOLUTION

Solution : A

Given that the sum of the distances from (x,y) to the points (2,0) and (-2,0) is 8.
Let the equation of the ellipse be $\frac{X^{2}}{a^{2}} + \frac{Y^{2}}{b^{2}} = 1$
$∴ 2 a = 8 \Rightarrow a = 4; c = 2 b^{2} = a^{2} - c^{2} \Rightarrow b^{2} = 16 - 4 = 12$

The equation of the ellipse is $\frac{X^{2}}{16} + \frac{Y^{2}}{12} = 1$

If (x,3) lies on the ellipse, then,
$\frac{x^{2}}{16} + \frac{3^{2}}{12} = 1 \Rightarrow \frac{x^{2}}{16} = \frac{1}{4} \Rightarrow x^{2} = 4 x = 2 [only positive square root is considered]$

Free Conic Sections 02 Practice Test - 11th Grade - Commerce

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

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