Free Objective Test 01 Practice Test - 11th and 12th

Question 1

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 2

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 3

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 4

The co-ordinates of the extremities of the latus rectum of the parabola $5 y^{2} = 4 x$ are

$(\frac{1}{5} . \frac{2}{5}), (\frac{- 1}{5}, \frac{2}{5})$

$(\frac{1}{5} . \frac{2}{5}), (\frac{1}{5}, \frac{- 2}{5})$

$(\frac{1}{5} . \frac{4}{5}), (\frac{1}{5}, \frac{- 4}{5})$

None of these

SOLUTION

Solution : B

$y^{2} = 4. \frac{1}{5} x; a = \frac{1}{5}$ . Focus is $(\frac{1}{5}, 0)$ and co-ordinates of latus rectum are $y^{2} = \frac{4}{25} \Rightarrow y = \pm \frac{2}{5}$

or end points of latus rectum are $(\frac{1}{5}, \pm \frac{2}{5})$ .

Question 5

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 6

The points on the parabola $y^{2} = 12 x$ whose focal distance is 4 , are

$(2, \sqrt{3}), (2, - \sqrt{3})$

$(\frac{1}{5}, \frac{2}{5}), (\frac{1}{5}, \frac{- 2}{5})$

(1, 2)

None of these

SOLUTION

Solution : D

a = 3 $\Rightarrow$ abscissa is 4 - 3 = 1 and $y^{2} = 12, y = \pm 2 \sqrt{3} .$

Hence points are $(1, 2 \sqrt{3}), (1, - 2 \sqrt{3})$ .

Question 7

A parabola passing through the point (-4,-2) has its vertex at the origin and y-axis as its axis. The latus rectum of

the parabola is

SOLUTION

Solution : B

Let the equation of parabola is $x^{2} = 4 a y,$ but a = $\frac{4}{- 2} = - 2. T h e n e q u a t i o n i s x^{2} = - 8 y$ and latus rectum = 4a = 8

Question 8

The lats rectum of a parabola whose directrix is x + y - 2 = 0 and focus is (3,-4), is

- $3 \sqrt{2}$

$3 \sqrt{2}$

$\frac{- 3}{\sqrt{2}}$

$\frac{3}{\sqrt{2}}$

SOLUTION

Solution : B

Distance between focus and directrix is

= $∣ ∣ ∣ \frac{3 - 4 - 2}{\sqrt{2}} ∣ ∣ ∣ = \frac{\pm 3}{\sqrt{2}}$

Hence latus rectum = $3 \sqrt{2}$

( Since latus rectum is two times the distance between focus and directrix ) .

Question 9

PQ is a double ordinate of the parabola $y^{2} = 4 a x$ . The locus of the points of trisection of PQ is

$9 y^{2} = 4 a x$

$9 x^{2} = 4 a y$

$9 y^{2} + 4 a x = 0$

$9 x^{2} + 4 a y = 0$

SOLUTION

Solution : A

Required locus is $(3 y)^{2}$ = 4ax

$\Rightarrow$ $9 y^{2} = 4 a x$ .

Question 10

If the parabola $y^{2} = 4 a x$ passes through (-3,2), then length of its latus rectum is

$\frac{2}{3}$

$\frac{1}{3}$

$\frac{4}{3}$

SOLUTION

Solution : C

The point (-3,2) will satisfy the equation $y^{2} = 4 a x$

$\Rightarrow$ 4 = -12a $\Rightarrow 4 a = - \frac{4}{3} = \frac{4}{3},$ (Taking positive sign).

Question 11

If the vertex of a parabola be at origin and directrix be x+5 = 0 , then its latus rectum is

SOLUTION

Solution : C

Distance between vertex and directrix = a = 5 units.
Therefore , latus rectum = 4a = 20

Question 12

If (2, 0) is the vertex and y-axis the directrix of a parabola, then its focus is

(2, 0)

(-2, 0)

(4, 0)

(-4, 0)

SOLUTION

Solution : C

Vertex = (2,0) $\Rightarrow$ focus is (2+2 ,0) = (4,0).

Question 13

The equation of the lines joining the vertex of the parabola $y^{2} = 6 x$ to the point on it whose abscissa is 24 , is

y ± 2x = 0

2y ± x = 0

x ± 2y = 0

2x ± y = 0

SOLUTION

Solution : B and C

$y^{2} = 6.24 \Rightarrow y = \pm 12$
So the equation of the lines is y=± 2x.

Question 14

The equation of the parabola with its vertex at the origin, axis on the y-axis and passing through the point (6, -3) is

$y^{3} = 12 x + 6$

$x^{2} = 12 y$

$x^{2} = - 12 y$

$y^{2} = - 12 x + 6$

SOLUTION

Solution : C

Since the axis of parabola is y-axis

$∴$ Equation of parabola $x^{2} = 4 a y$

Since it passes through (6 , -3)

$∴$ 36 = -12 a $\Rightarrow a = - 3$
$∴$ Equation of parabola is $x^{2} = - 12 y$ .

Question 15

Two tangents are drawn from the point $(- 2, - 1)$ to the parabola $y^{2} = 4 x$ . If $α$ is the angle between those tangents then tan $α$ =

3

\frac{1}{3}

C. 2

\frac{1}{2}

SOLUTION

Solution : A

The equation to the pair of tangents is $[- y - 2 (x - 2)]^{2} = (1 + 8) (y^{2} - 4 x) \Rightarrow (- 2 x - y + 4)^{2} = - (y^{2} - 4 x)$
$\Rightarrow 4 x^{2} + y^{2} + 16 - 16 x - 8 y + 4 x y = 9 y^{2} - 36 x \Rightarrow 4 x^{2} + 4 x y - 8 y^{2} + 20 x - 8 y + 16 = 0$
$∴ t a n α = \frac{2 \sqrt{4 + 32}}{| 4 - 8 |} = \frac{2 \times 6}{4} = 3$

Free Objective Test 01 Practice Test - 11th and 12th

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

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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