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