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

### Question 1

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

#### 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)

(h−ae2)2(a2)2+k2(b2)2=1

Area=πab4

Ratio=14

### Question 2

If the ellipse x24+y21=1 meet the ellipse x21+y2a2=1 in four distinct points and a=b2−10b+25, then the value b does not satisfy

#### SOLUTION

Solution :D

For the two ellipse to intersect at four different points

1<a2⇒1<(b−5)2⇒b/ϵ[4,6]

### Question 3

If the ellipse x2a2−3+y2a+4=1 is inscribed in a square of side length a√2 then a is

#### 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√(a2−3)+(a+4)=√2a2+2a22√a2+a+1=2a⇒a=−1But for ellipse a2>3&a>−4

So there is no value of a which satisfy both.

### Question 4

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

#### SOLUTION

Solution :D

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

Area=π ab

### Question 5

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

#### 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+b2tan2θ∴Minimum value of a2cot2θ+b2tan2θ is 2ab(∵A.M≥G.M)∴Minimum value of MN=a+b

### Question 6

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

#### SOLUTION

Solution :D

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

### Question 7

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 :

#### 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

### Question 8

The line lx + my + n = 0 will be a normal to the hyperbola b2x2−a2y2=a2b2, if

#### SOLUTION

Solution :D

The equation of the normal at (asecϕ,btanϕ) to the hyperbola x2a2−y2b2=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)tanϕ−n=(a2+b2)sinϕ−ncosϕ

∴ sinϕ=blamand cosϕ=(a2+b2)l−na

⇒ a2l2−b2m2=(a2+b2)2−n2

### Question 9

Equation of the chord of the hyperbola 25x2−16y2=400 which is bisected at the point (6, 2), is

#### SOLUTION

Solution :B

Equation of hyperbola is 25x2−16y2=400

⇒ x216−y225=1

Chord is bisected at (6, 2)

∴ 6x16−2y25=6216−2225

⇒ 75x−16y=418

### Question 10

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

#### 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+2fgh−af2−bg2−ch2=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)]

### Question 11

The eccentricity of the rectangular hyperbola is

#### SOLUTION

Solution :B

In case of rectangular hyperbola,

a=b⇒b2=a2⇒a2(e2−1)=a2⇒e=√2

### Question 12

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

#### SOLUTION

Solution :B

Let the equation to the rectangular hyperbola be

x2−y2=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

x−ysinα=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.

### Question 13

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

#### 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+2y2−5xy+10=0

### Question 14

If two tangents can be drawn to the different branches of hyperbola

x21−y24=1 from the paint (α,α2), then

#### SOLUTION

Solution :C

Given that,

x21−y24=1

Since, (α,α2) lie on the parabola y=x2, then (α,α2) must lie between the asymptotes of hyperbola x21−y24=1 in 1st and 2nd quadrants.

So, the asymptotes are y=±2x

∴ 2α<α2

⇒ α<0orα>2 and −2α<α2

α<−2or−2α<0

∴ αϵ(−∞,−2)or(2,∞)

### Question 15

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

#### SOLUTION

Solution :D

ransverse axis is the equation of the angle bisector passing containing point (2, 3), which is given by

3x−4y+55=12x+5y−4013

⇒ 21x+77y=265