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

### Question 1

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

#### 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+b2⇒4a2+b2=16

using A.M.≥G.M.,We get ab≤4

### Question 2

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

#### 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

y−b2a=1e(x−ae)

It passes through´B(0,b) then

b−b2a=−a

a2−b2=−ab

a4e4=a2b2

e4+e2=1

### Question 3

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

#### SOLUTION

Solution :A

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

### Question 4

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

#### SOLUTION

Solution :A

1SP+1SQ=2ab2

a= 5

b=4

Solving we get,

SQ=2.

### Question 5

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

#### SOLUTION

Solution :D

x2a2+y2b2=1,Area=πabLet P=(acosθ,bsinθ)S=(ae,0)M(h,k)midpoint of PS⇒h=ae+acosθ2;k=bsinθ2⇒(h−ae2a2)2+(kb2)2=1

Area = 14πab

### Question 6

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

#### SOLUTION

Solution :B

Equation of tangent at a point (acosθ,bsinθ)isxacosθ+ybsinθ=1

But, it is the same as xa√32+yb.12=1

∴cosθ=√32,sinθ=12⇒θ=π6

### Question 7

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

#### 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+2yy1−4=0

This touches circle ⇒r2(l2+m2)=n2⇒1(x21+4y21)=16⇒x2+4y2=16 is ellipse with eccentricity e=√32 and length of latus rectum LL1=2

### Question 8

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

#### SOLUTION

Solution :D

Equation of normal at p(3cosθ,2sinθ) is 3xsecθ−2ycosecθ=5

5√9sec2θ+4cosec2θ=√3But minimum value of 9sec2θ+4cosec2θ=25∴no such θ−1 exists

So the number of tangents to the circle x2+y2=3 which are normal to the ellipse x29+y24=1 is 0.

### Question 9

The equation of a tangent parallel to y = x drawn to x23−y22=1, is

#### SOLUTION

Solution :A

Let y = x + c be a tangent to x23−y22=1, then c2=3−2=1 ⇒ c=±1 So, the required tangents are y=x±1

### Question 10

The locus of the middle points of chords of hyperbola 3x2−2y2+4x−6y=0 parallel to y = 2x, is

#### SOLUTION

Solution :A

By T = S1, the equation of chord whose mid-point is (α,β) is 3xα−2yβ+(x+α)−2(y+β)=0

⇒ x(3α+2)−y(2β+3)=2

as it is parallel to y = 2x

∴ 3α–4β=4

∴ Required locus is 3x – 4y = 4

### Question 11

If x = 9 is the chord of contact of the hyperbola x2−y2=9, then the equation of the corresponding pair of tangents is

#### SOLUTION

Solution :B

If x = 9 meets the hyperbola at (9,6√2) and (9,−6√2).

Then, equations of tangent at these points are 3x−2√2y−3=0 and 3x+2√2y−3=0.

The combined equation of these two is 9x2−8y2−18x+9=0.

### Question 12

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

#### 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.

### Question 13

From any point on the hyperbola x2a2−y2b2=1, tangents are drawn to the hyperbola x2a2−y2b2=2 . Then, area cut-off by the chord of contact on the asymptotes is equal to

#### 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 xa−yb=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(x1x2−x2y1)

=12∣∣ ∣ ∣∣⎛⎜⎝−4ab×2x21a2−y21b2⎞⎟⎠∣∣ ∣ ∣∣

### Question 14

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

r2

2r2

3r2

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

x4−r2x2+c4=0Now,CP2+CQ2+CR2+CS2

=x21+y21+x22+y22+x23+y23+x24+y24

=(∑4i=1xi)2−2∑x1x2+=(∑4i=1yi)2−2∑y1y2

=2r2+2r2=4r2 [from equation (iii)]

### Question 15

The product of the perpendicular from any point on the hyperbola x2a2−y2b2=1 to its asymptotes, is equal to

#### SOLUTION

Solution :B

Let (a secθ,b tanθ) be the any point on the hyperbola x2a2−y2b2=1

The equations of the asymptotes of the given hyperbola are xa−yb=0 and xa+yb=0

Now, p1 = length of the perpendicular from (a secθ,b tanθ) on xa+yb=0

=secθ−tanθ√1a2+1b2

and p2 = length of the perpendicular from (a secθ,btanθ) xa+yb=0

=secθ−tanθ√1a2+1b2

∴ p1p2=sec2θ−tan2θ1a2+1b2=a2b2a2+b2