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

If a tangent of slope 2 of the ellipse x2a2+y2b2=1 is normal to the circle x2+y2+4x+1=0 then the maximum value of ab is

A. 2
B. 4
C. 6
D. Can’t be found

#### SOLUTION

Solution : B

A tangent of slope 2 is y=2x±4a2+b2 ,this is normal to x2+y2+4x+1=0
then 0=4±4a2+b24a2+b2=16
using A.M.G.M.,We get ab4

The normal at an end of a latus rectum of the ellipse x2a2+y2b2=1 passes through an end of the minor  axis if

A. e4+e2=1
B. e3+e2=1
C. e2+e=1
D. e3+e=1

#### SOLUTION

Solution : A

Given ellipse equation is x2a2+y2b2=1
Let P(ae,b2a) be one end of latus rectum.
Slope of normal at P(ae,b2a)=1e
Equation of normal is
yb2a=1e(xae)
It passes through´B(0,b) then
bb2a=a
a2b2=ab
a4e4=a2b2
e4+e2=1

Consider two points A and B on the ellipse x225+y29=1
circles are drawn having segments of tangents at A and B in between tangents at the two ends of major axis of ellipse as diameter, then the length of common chord of the circles is

A. 8
B. 6
C. 10
D. 42

#### SOLUTION

Solution : A

All such circles pass through foci  The common  chord is of the length 2ae = 10×45=8

If  PSQ is a focal chord of the ellipse 16x2+25y2=400 such that SP = 8 then the length of  SQ =

A. 2
B. 113
C. 16
D. 25

#### SOLUTION

Solution : A

1SP+1SQ=2ab2
a= 5
b=4
Solving we get,
SQ=2.

The ratio of the area enclosed by the locus of mid-point of PS and area of the ellipse where P is any point on the ellipse and S is the focus of the ellipse, is

A. 12
B. 13
C. 15
D. 14

#### SOLUTION

Solution : D

x2a2+y2b2=1,Area=πabLet P=(acosθ,bsinθ)S=(ae,0)M(h,k)midpoint of PSh=ae+acosθ2;k=bsinθ2(hae2a2)2+(kb2)2=1
Area = 14πab

If (3)bx+ay=2ab is tangent to the ellipse x2a2+y2b2=1 , then eccentric angle θ is

A. π4
B. π6
C. π2
D. π3

#### SOLUTION

Solution : B

Equation of tangent at a point (acosθ,bsinθ)isxacosθ+ybsinθ=1
But, it is the same as xa32+yb.12=1
cosθ=32,sinθ=12θ=π6

If a variable tangent of the circle x2+y2=1 intersect the ellipse x2+2y2=4 at P and Q then the locus of the points of intersection of the tangents at P and Q is

A.   A circle of radius 2 units
B.   A parabola with fouc as (2, 3)
C. An ellipse with eccentricity 34
D. An ellipse with length of latus rectrum is 2 units

#### SOLUTION

Solution : D

x2+y2=1; x2+2y2=4
Let R(x1,y1) is point of intersection of tangents drawn at P,Q to ellipse PQ is chord of contact of R(x1,y1)
xx1+2yy14=0
This touches circle r2(l2+m2)=n21(x21+4y21)=16x2+4y2=16 is ellipse with eccentricity e=32 and length of latus rectum LL1=2

How many tangents to the circle x2+y2=3 are there which are normal to the ellipse x29+y24=1

A. 3
B. 2
C. 1
D. 0

#### SOLUTION

Solution : D

Equation of normal at p(3cosθ,2sinθ) is 3xsecθ2ycosecθ=5
59sec2θ+4cosec2θ=3But minimum value of 9sec2θ+4cosec2θ=25no such θ1 exists
So the number of tangents to the circle x2+y2=3 which are normal to the ellipse x29+y24=1 is 0.

The equation of a tangent parallel to y = x drawn to x23y22=1, is

A.  x -y + 1 = 0
B. x - y + 2 = 0
C. x + y - 1 = 0
D. x + y – 2 = 0

#### SOLUTION

Solution : A

Let y = x + c be a tangent to x23y22=1, then c2=32=1      c=±1    So, the required tangents are y=x±1

The locus of the middle points of chords of hyperbola 3x22y2+4x6y=0  parallel to y = 2x, is

A. 3x – 4y = 4
B. 3y – 4x + 4 = 0
C. 4x – 4y = 3
D. 3x – 4y = 2

#### SOLUTION

Solution : A

By T = S1, the equation of chord whose mid-point is (α,β) is 3xα2yβ+(x+α)2(y+β)=0
x(3α+2)y(2β+3)=2
as it is parallel to y = 2x
3α4β=4
Required locus is 3x – 4y = 4

If  x = 9 is the chord of contact of the hyperbola x2y2=9, then the equation of the corresponding pair of tangents is

A. 9x28y2+18x9=0
B. 9x28y218x+9=0
C. 9x28y218x9=0
D. 9x28y2+18x+9=0

#### SOLUTION

Solution : B

If x = 9 meets the hyperbola at (9,62) and (9,62).
Then, equations of tangent at these points are 3x22y3=0 and 3x+22y3=0.
The combined equation of these two is 9x28y218x+9=0.

The locus of the point which is such that the chord of contact of tangents drawn from it to the ellipse x2a2+y2b2=1 forms a triangle of constant area with the coordinate axes, is

A. A straight line
B. A hyperbola
C. An ellipse
D. A circle

#### SOLUTION

Solution : B

The chord of contact of tangents from (x1y1)isxx1a2+yy1b2=1
It meets the axes at the points (a2x,0) and (0,b2y1)
Area of the triangle is 12.a2x1.b2y1 [constant]
x1y1=a2b22k=c2 where c is a constant.
xy=c2 is the required locus.

From any point on the hyperbola x2a2y2b2=1, tangents are drawn to the hyperbola x2a2y2b2=2 . Then, area  cut-off by the chord of contact on the asymptotes is equal to

A. a/2 sq unit
B. ab sq unit
C. 2ab sq unit
D. 4ab sq unit

#### SOLUTION

Solution : D

Let P(x1,y1) be a point on the hyperbola x2a2+y2b2=1
The chord of contact of tangents from P to the hyperbola is given by  xx1a2+yy1b2=1   …… (i)
The equation of the asymptotes are xayb=0
and  xa+yb=0
The points of intersection of Equation (i) with the two asymptotes are given by
x1=2ax1a+y1b,y1=2ax1a+y1b
x2=2ax1a+y1b,y2=2ax1a+y1b
Area of the triangle = 12(x1x2x2y1)
=12∣ ∣ ∣4ab×2x21a2y21b2∣ ∣ ∣

A rectangular hyperbola whose centre is C, is cut by any circle of radius r in four points P, Q, R and S. Then, CP2+CQ2+CR2+CS2 is equal to

A.

r2

B.

2r2

C.

3r2

D.

4r2

#### SOLUTION

Solution : D

Let equation of the rectangular hyperbola be xy=c2 and equation of circle be x2+y2=r2
Put y=c2x in equation (ii), we get x2+c4x2r2
x4r2x2+c4=0Now,CP2+CQ2+CR2+CS2
=x21+y21+x22+y22+x23+y23+x24+y24
=(4i=1xi)22x1x2+=(4i=1yi)22y1y2
=2r2+2r2=4r2 [from equation (iii)]

The product of the perpendicular from any point on the hyperbola x2a2y2b2=1 to its asymptotes, is equal to

A. aba+b
B. a2b2a2+b2
C. 2a2b2a2+b2
D. None of these

#### SOLUTION

Solution : B

Let (a secθ,b tanθ) be the any point on the hyperbola x2a2y2b2=1
The equations of the asymptotes of the given hyperbola are xayb=0 and xa+yb=0
Now, p1 = length of the perpendicular from (a secθ,b tanθ) on xa+yb=0
=secθtanθ1a2+1b2
and p2 = length of the perpendicular from (a secθ,btanθ) xa+yb=0
=secθtanθ1a2+1b2
p1p2=sec2θtan2θ1a2+1b2=a2b2a2+b2