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

The end points of latus rectum of the parabola x2=4ay are

A.

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

B.

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

C.

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

D.

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

#### SOLUTION

Solution : D

It is a fundamental concept. The end points of latus rectum of the parabola x2=4ay are (-2a , a) , (2a, a).

The ends of latus rectum of parabola x2+8y=0 are

A.

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

B.

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

C.

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

D.

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

#### SOLUTION

Solution : C

x2=8ya=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.

The focus of the parabols x2=16y is

A.

(4, 0)

B.

(0, 4)

C.

(-4, 0)

D.

(0, -4)

#### SOLUTION

Solution : D

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

The co-ordinates of the extremities of the latus rectum of the parabola 5y2=4x are

A.

(15.25),(15,25)

B.

(15.25),(15,25)

C.

(15.45),(15,45)

D.

None of these

#### SOLUTION

Solution : B

y2=4.15x;a=15 . Focus is (15,0) and co-ordinates of latus rectum are y2=425y=±25

or end points of latus rectum are (15,±25)

The points on the parabola y2=36x whose ordinate is three times the absicca are

A.

(0, 0), (4, 12)

B.

(1, 3), (4, 12)

C.

(4, 12)

D.

None of these

#### SOLUTION

Solution : A

y1=3x1 . According to given condition 9x21=36x1

x1=4,0y1=12,0

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

The points on the parabola y2=12x whose focal distance is 4 , are

A.

(2,3),(2,3)

B.

(15,25),(15,25)

C.

(1, 2)

D.

None of these

#### SOLUTION

Solution : D

a = 3 abscissa is 4 - 3 = 1 and y2=12,y=±23.

Hence points are (1,23),(1,23)

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

A.

6

B.

8

C.

10

D.

12

#### SOLUTION

Solution : B

Let the equation of parabola is x2=4ay, but a = 42=2.Then equation is x2=8y and latus rectum = 4a = 8

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

A.

32

B.

32

C.

32

D.

32

#### SOLUTION

Solution : B

Distance between focus and directrix is

= 3422=±32

Hence latus rectum = 32

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

PQ is a double ordinate of the parabola y2=4ax . The locus of the points of trisection of PQ is

A.

9y2=4ax

B.

9x2=4ay

C.

9y2+4ax=0

D.

9x2+4ay=0

#### SOLUTION

Solution : A

Required locus is (3y)2 = 4ax

9y2=4ax

If the parabola y2=4ax passes through (-3,2), then length of its latus rectum is

A.

23

B.

13

C.

43

D.

4

#### SOLUTION

Solution : C

The point (-3,2) will satisfy the equation y2=4ax

4 = -12a 4a=43=43, (Taking positive sign).

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

A.

5

B.

10

C.

20

D.

40

#### SOLUTION

Solution : C

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

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

A.

(2, 0)

B.

(­-2, 0)

C.

(4, 0)

D.

(-4, 0)

#### SOLUTION

Solution : C

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

The equation of the lines joining the vertex of the parabola y2=6x  to the point on it whose abscissa is 24 , is

A.

y ± 2x = 0

B.

2y ± x = 0

C.

x ± 2y = 0

D.

2x ± y = 0

#### SOLUTION

Solution : B and C

y2=6.24y=±12
So the equation of the lines is y=± 2x.

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

A.

y3=12x+6

B.

x2=12y

C.

x2=12y

D.

y2=12x+6

#### SOLUTION

Solution : C

Since the axis of parabola is y-axis

Equation of parabola x2=4ay

Since it passes through (6 , -3)

36 = -12 a a=3

Equation of parabola is x2=12y .

Two tangents are drawn from the point (2,1)  to the parabola y2=4x. If α  is the angle between those tangents then tan α=

A. 3
B. 13
C. 2
D. 12

#### SOLUTION

Solution : A

The equation to the pair of tangents is [y2(x2)]2=(1+8)(y24x)(2xy+4)2=(y24x)
4x2+y2+1616x8y+4xy=9y236x4x2+4xy8y2+20x8y+16=0
tanα=24+32|48|=2×64=3