# Free Objective Test 02 Practice Test - 11th and 12th

The area of the parallelogram formed by the tangents at the points whose eccentric angles are θ,θ+π2,θ+π,θ+3π2 on the ellipse x2a2+y2b2=1 is

A. ab
B. 4ab
C. 3ab
D. 2ab

#### SOLUTION

Solution : B x2a2+y2b2=1
Area=πab
Let P =(a cosθ,b sinθ)
S=(ae,0)
M(h,k) be the midpoint of PS
(h,k)=(ae+a cosθ2,b sinθ2)
(hae2)2(a2)2+k2(b2)2=1
Area=πab4
Ratio=14

If the ellipse x24+y21=1 meet the ellipse x21+y2a2=1 in four distinct points and a=b210b+25, then the value b does not satisfy

A. (,4)
B. (4, 6)
C. (6,)
D. [4, 6]

#### SOLUTION

Solution : D

For the two ellipse to intersect at four different points
1<a21<(b5)2b/ϵ[4,6]

If the ellipse x2a23+y2a+4=1 is inscribed in a square of side length a2 then a is

A. 4
B. 2
C. 1
D. None of these

#### SOLUTION

Solution : D

Sides of the square will be perpendicular tangents to the ellipse so, vertices of the square will lie on director circle. So diameter of director circle is
2(a23)+(a+4)=2a2+2a22a2+a+1=2aa=1But for ellipse a2>3&a>4
So there is no value of a which satisfy both.

A man running round a race course notes that the sum of the distances of two flag posts from him is 8 meters.The area of the path he encloses in square meters if the distance between flag posts is 4 is

A. 153π
B. 123π
C. 183π
D. 83π

#### SOLUTION

Solution : D

The path he runs is an ellipse whose major axis is 4m,e=12
Area=π ab

A tangent to the ellipse x2a2+y2b2=1 cuts the axes in M and N. Then the least length of MN is

A. a + b
B. a - b
C. a2+b2
D. a2b2

#### SOLUTION

Solution : A

Equation of tangent at (θ) is
xacosθ+ybsinθ=1It meets axes at M=(acosθ,0)and N=(0,bsin θ)MN=a2sec2θ+b2cosec2θ=a2+b2+a2cot2θ+b2tan2θMinimum value of a2cot2θ+b2tan2θ is 2ab(A.MG.M)Minimum value of MN=a+b

Tangents are drawn from any point on the circle x2+y2=41 to the Ellipse x225+y216=1 then the angle between the two tangents is

A. π4
B. π3
C. π6
D. π2

#### SOLUTION

Solution : D

The given circle is the director circle of the ellipse,so the angle between the tangents is π2.

The perimeter of a triangle is 20 and the points (-2, -3) and (-2, 3) are two of the vertices of it. Then the locus of third vertex is :

A. (x2)249+y240=1
B. (x+2)249+y240=1
C. (x+2)240+y249=1
D. (x2)240+y249=1

#### SOLUTION

Solution : C

A(2,3),B(2,3)
Let the third vertex be C(x,y)
AB + BC+CA=20
BC+CA=14
So locus of C is ellipse with A,B as focii.
2a=14
2ae=6
e=37
we get b as 40
Equation of ellipse is (x+2)240+y249=1

The line lx + my + n = 0 will be a normal to the hyperbola b2x2a2y2=a2b2, if

A. a2l2+b2m2=(a2+b2)2n2

B. a2l2+b2m2=(a2b2)2n2

C. a2l2+b2m2=(a2+b2)2n
D. None of these

#### SOLUTION

Solution : D

The equation of the normal at (asecϕ,btanϕ) to the hyperbola  x2a2y2b2=1 is
ax sinϕ+by=(a2+b2)tanϕ     ......(i)
and the equation of the line is
lx + my + n =0                            …… (ii)
now, equations (i) and (ii) represent the same line, therefore
2sinϕl=bm=(a2+b2)tanϕn=(a2+b2)sinϕncosϕ
sinϕ=blamand cosϕ=(a2+b2)lna
a2l2b2m2=(a2+b2)2n2

Equation of the chord of the hyperbola 25x216y2=400 which is bisected at the point (6, 2), is

A. 16x – 75y = 418
B. 75x - 16y = 418
C. 25x - 4y = 400
D. None of these

#### SOLUTION

Solution : B

Equation of hyperbola is 25x216y2=400
x216y225=1
Chord is bisected at (6, 2)
6x162y25=62162225
75x16y=418

The equation of a hyperbola, conjugate to the hyperbola x2+3xy+2y2+2x+3y+1=0, is

A.  x2+3xy+2y2+2x+3y+1=0
B. x2+3xy+2y2+2x+3y+2=0
C.  x2+3xy+2y2+2x+3y+3=0
D. x2+3xy+2y2+2x+3y+4=0

#### SOLUTION

Solution : B

We know that,
Equation of hyperbola + Equation of conjugate
hyperbola = 2(Equation of asymptotes)
Equation of conjugate hyperbola
= 2 Equation of asymptotes - Equation of hyperbola
C(x, y) = 2A(x, y) – H(x, y)           …… (i)
H(x,y)x2+3xy+2y2+2x+3y=0
Equation of asymptotes is
x2+3xy+2y2+2x+3y+λ=0
Δ=0,abc+2fghaf2bg2ch2=0,thenλ=1
A(x,y)x2+3xy+2y2+2x+3y+1=0
C(x,y)x2+3xy+2y2+2x+3y+2=0[from equation (i)]

The eccentricity of the rectangular hyperbola is

A. 2
B. 2
C. 0
D. None of these

#### SOLUTION

Solution : B

In case of rectangular hyperbola,
a=bb2=a2a2(e21)=a2e=2

Two concentric hyperbolas,whose axes meet at angle of 45,cut

A. at 450
B. at 900
C. Nothing can be said
D. None of these

#### SOLUTION

Solution : B

Let the equation to the rectangular hyperbola be
x2y2=a2       …… (i)
As the asymptotes of this are the axes of the other and vice-versa, hence the equation of the other hyperbola may be written as
xy=c2         …… (ii)
Let equations (i) and (ii) meet at some point whose coordinates are
(a secα,a tanα)
Then, the tangent at the point a (secα,atanα) to equation (ii) is
xysinα=acosα     …… (iii)
and the tangent at the point (a secα,atanα) to equation (ii) is
y+xsinα=(2c2a)cosα           …… (iv)
clearly, the slopes of the tangents given by equations (iv) and (iii) are respectively - secα and1sinα , so their product is secα.(1sinα)=1 Hence, the tangents are at right angle.

If foci of hyperbola lie on y = x and one of the asymptotes is y = 2x, then equation of the hyperbola, given  that it passes through (3, 4), is

A. x2y252xy+5=0
B. 2x22y2+5xy+5=0
C. 2x2+2y25xy+10=0
D.  None of these

#### SOLUTION

Solution : C

Foci of hyperbola lie on y = x. So, the major axis is y = x. Major axis of hyperbola bisects the asymptote.
Equation of hyperbola is x = 2y
Equation of hyperbola is (y – 2x)(x – 2y) + k = 0
Given that, it passes through (3, 4)
Hence, required equation is
2x2+2y25xy+10=0

If two tangents can be drawn to the different branches of hyperbola
x21y24=1 from the paint (α,α2),        then

A. αϵ(2,0)

B. αϵ(3,0)
C. αϵ(,2)

D. αϵ(,3)

#### SOLUTION

Solution : C

Given that,
x21y24=1 Since, (α,α2) lie on the parabola y=x2, then (α,α2) must lie between the asymptotes of hyperbola  x21y24=1 in 1st and 2nd quadrants.
So, the asymptotes are y=±2x
2α<α2
α<0orα>2 and 2α<α2
α<2or2α<0
αϵ(,2)or(2,)

If a hyperbola passes through (2, 3) and has asymptotes 3x - 4y + 5 = 0 and 12x + 5y - 40 = 0, then the equation of its transverse axis is

A. 77x - 21y - 265 = 0
B. 21x - 77y + 265 = 0
C. 21x - 77y - 265 = 0
D. 21x + 77y - 265 = 0

#### SOLUTION

Solution : D

ransverse axis is the equation of the angle bisector passing containing point (2, 3), which is given by
3x4y+55=12x+5y4013
21x+77y=265