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Hyperbola

JEE Mathematics · 61 questions · Page 1 of 7 · Click an option or "Show Solution" to reveal answer

Q1
The eccentricity of the hyperbola whose length of the latus rectum is equal to 88 and the length of its conjugate axis is equal to half of the distance between its foci, is :
A 23{2 \over {\sqrt 3 }}
B 3{\sqrt 3 }
C 43{{4 \over 3}}
D 43{4 \over {\sqrt 3 }}
Correct Answer
Option A
Solution
2b2a=8{{2{b^2}} \over a} = 8

and

2b=12(2ae)2b = {1 \over 2}\left( {2ae} \right)
⇒4b2=a2e2\Rightarrow 4{b^2} = {a^2}{e^2}
⇒4a2(e2−1)=a2e2\Rightarrow 4{a^2}\left( {{e^2} - 1} \right) = {a^2}{e^2}
⇒3e2=4⇒e=23\Rightarrow 3{e^2} = 4 \Rightarrow e = {2 \over {\sqrt 3 }}
Q2
If a hyperbola passes through the point P(10, 16) and it has vertices at (± 6, 0), then the equation of the normal to it at P is :
A 2x + 5y = 100
B x + 3y = 58
C x + 2y = 42
D 3x + 4y = 94
Correct Answer
Option A
Solution

Let hyperbola is

x2a2−y2b2=1{{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1

vertices (±\pma, 0 ) = (±\pm6, 0) ⇒\Rightarrow a = 6 Hyperbola passes through p(10, 16) ∴\therefore

10262−162b2=1{{{{10}^2}} \over {{6^2}}} - {{{{16}^2}} \over {{b^2}}} = 1

⇒\Rightarrow b = 12 ∴\therefore Required hyperbola is

x236−y2144=1{{{x^2}} \over {36}} - {{{y^2}} \over {144}} = 1

Equation of normal will be

a2xx1+b2yy1=a2+b2{{{a^2}x} \over {{x_1}}} + {{{b^2}y} \over {{y_1}}} = {a^2} + {b^2}

At P(10,16) normal is

36x10+144y16=36+144{{36x} \over {10}} + {{144y} \over {16}} = 36 + 144

⇒\Rightarrow 18x + 45y = 900 ⇒\Rightarrow 2x + 5y = 100

Q3
The normal to a curve at P(x,y)P(x,y) meets the xx-axis at GG. If the distance of GG from the origin is twice the abscissa of PP, then the curve is a :
A circle
B hyperbola
C ellipse
D parabola
Correct Answer
Option B
Solution

Equation of normal at

P(x,y)P\left( {x,y} \right)

is

Y−y=−dxdy(x−x)Y - y = - {{dx} \over {dy}}\left( {x - x} \right)

Coordinate of

GG

at

XX

axis is

(X,0)\left( {X,0} \right)

(let) ∴\therefore

0−y=−dxdy(X−x)⇒ydydx=X−x0 - y = - {{dx} \over {dy}}\left( {X - x} \right) \Rightarrow y{{dy} \over {dx}} = X - x
⇒X=x+ydydx\Rightarrow X = x + y{{dy} \over {dx}}

∴\therefore Co-ordinate of

G(x+ydydx,0)G\left( {x + y{{dy} \over {dx}},0} \right)

Given distance of

GG

from origin == twice of the abscissa of

P.P.

as distance cannot be

−ve,-ve,

therefore abscissa

xx

should be

+ve+ve

∴\therefore

x+ydydx=2x⇒ydydx=x⇒ydx=xdxx + y{{dy} \over {dx}} = 2x \Rightarrow y{{dy} \over {dx}} = x \Rightarrow ydx = xdx

On Integrating

⇒y22=x22+c1⇒x2−y2=−2c1\Rightarrow {{{y^2}} \over 2} = {{{x^2}} \over 2} + {c_1} \Rightarrow {x^2} - {y^2} = - 2{c_1}

∴\therefore the curve is a hyperbola

Q4
The foci of the ellipse x216+y2b2=1{{{x^2}} \over {16}} + {{{y^2}} \over {{b^2}}} = 1 and the hyperbola x2144−y281=125{{{x^2}} \over {144}} - {{{y^2}} \over {81}} = {1 \over {25}} coincide. Then the value of b2{b^2} is :
A 99
B 11
C 55
D 77
Correct Answer
Option D
Solution
x2144−y281=125{{{x^2}} \over {144}} - {{{y^2}} \over {81}} = {1 \over {25}}
a=14425,b=8125,  a = \sqrt {{{144} \over {25}}} ,b = \sqrt {{{81} \over {25}}} ,\,\,
e=1+81144=1512=54e = \sqrt {1 + {{81} \over {144}}} = {{15} \over {12}} = {5 \over 4}

∴\therefore Foci

=(±3,0)= \left( { \pm 3,0} \right)

∴\therefore foci of ellipse == foci of hyperbola ∴\therefore for ellipse

ae=3ae=3

but

a=4,a=4,

∴\therefore

e=34e = {3 \over 4}

Then

b2=a2(1−e2){b^2} = {a^2}\left( {1 - {e^2}} \right)
⇒b2=16(1−916)=7\Rightarrow {b^2} = 16\left( {1 - {9 \over {16}}} \right) = 7
Q5
The locus of a point P(α,β)P\left( {\alpha ,\beta } \right) moving under the condition that the line y=αx+βy = \alpha x + \beta is tangent to the hyperbola x2a2−y2b2=1{{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1 is :
A an ellipse
B a circle
C a parabola
D a hyperbola
Correct Answer
Option D
Solution

Tangent to the hyperbola

x2a2−y2b2=1{{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1

is

y=mx±a2m2−b2y = mx \pm \sqrt {{a^2}{m^2} - {b^2}}

Given that

y=αx+βy = \alpha x + \beta

is the tangent of hyperbola

⇒m=α\Rightarrow m = \alpha

and

a2m2−b2=β2{a^2}{m^2} - {b^2} = {\beta ^2}

∴\therefore

a2α2−b2=β2{a^2}{\alpha ^2} - {b^2} = {\beta ^2}

Locus is

a2x2−y2=b2{a^2}{x^2} - {y^2} = {b^2}

which is hyperbola.

Q6
For the Hyperbola x2cos⁡2α−y2sin⁡2α=1{{{x^2}} \over {{{\cos }^2}\alpha }} - {{{y^2}} \over {{{\sin }^2}\alpha }} = 1 , which of the following remains constant when α\alpha varies==?
A abscissae of vertices
B abscissae of foci
C eccentricity
D directrix.
Correct Answer
Option B
Solution

Given, equation of hyperbola is

x2cos⁡2α−y2sin⁡2α=1{{{x^2}} \over {{{\cos }^2}\alpha }} - {{{y^2}} \over {{{\sin }^2}\alpha }} = 1

We know that the equation of hyperbola is

x2a2−y2b2=1{{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1

Here,

a2=cos⁡2α{a^2} = {\cos ^2}\alpha

and

b2=sin⁡2α{b^2} = {\sin ^2}\alpha

We know that,

b2=a2(e2−1){b^2} = {a^2}\left( {{e^2} - 1} \right)
⇒sin⁡2α=cos⁡2α(e2−1)\Rightarrow {\sin ^2}\alpha = {\cos ^2}\alpha \left( {{e^2} - 1} \right)
⇒sin⁡2α+cos⁡2α=cos⁡2α.e2\Rightarrow {\sin ^2}\alpha + {\cos ^2}\alpha = {\cos ^2}\alpha .{e^2}
⇒e2=1+tan⁡2α=sec⁡2α⇒e=sec⁡ α\Rightarrow {e^2} = 1 + {\tan ^2}\alpha = {\sec ^2}\alpha \Rightarrow e = \sec \,\alpha

∴\therefore

ae=cos⁡α  .  1cos⁡α=1ae = \cos \alpha \,\,.\,\,{1 \over {\cos \alpha }} = 1

Co-ordinate of foci are

(±αe,0)  \left( { \pm \alpha e,0} \right)\,\,

i.e.

(±1,0)\left( { \pm 1,0} \right)

Hence, abscissae of foci remain constant when α\alpha varies.

Q7
A hyperbola passes through the point P(2,3)\left( {\sqrt 2 ,\sqrt 3 } \right) and has foci at (±2,0)\left( { \pm 2,0} \right). Then the tangent to this hyperbola at P also passes through the point :
A (22,33)\left( {2\sqrt 2 ,3\sqrt 3 } \right)
B (3,2)\left( {\sqrt 3 ,\sqrt 2 } \right)
C (−2,−3)\left( { - \sqrt 2 , - \sqrt 3 } \right)
D (32,23)\left( {3\sqrt 2 ,2\sqrt 3 } \right)
Correct Answer
Option A
Solution

Equation of hyperbola is

x2a2−y2b2=1{{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1

foci is (±2, 0) ⇒\Rightarrow ae = 2 ⇒\Rightarrow a2e2 = 4 Since b2 = a2 (e2 – 1) b2 = a2 e2 – a2 ∴\therefore a2 + b2 = 4 .....(

1) Also Hyperbola passes through

(2,3)\left( {\sqrt 2 ,\sqrt 3 } \right)

∴\therefore

2a2−3b2=1{2 \over {{a^2}}} - {3 \over {{b^2}}} = 1

⇒\Rightarrow

24−b2−3b2=1{2 \over {4 - {b^2}}} - {3 \over {{b^2}}} = 1

⇒\Rightarrow (b2 – 3) (b2 + 4) = 0 ∴\therefore b2 = 3 or b2 = -4 For b2 = 3 ⇒\Rightarrow a2 = 1

x21−y23=1{{{x^2}} \over 1} - {{{y^2}} \over 3} = 1

Equation of tangent is

2x1−3y3=1{{\sqrt 2 x} \over 1} - {{\sqrt 3 y} \over 3} = 1

It satisfy point

(22,33)\left( {2\sqrt 2 ,3\sqrt 3 } \right)

.

Q8
Tangents are drawn to the hyperbola 4x2 - y2 = 36 at the points P and Q. If these tangents intersect at the point T(0, 3) then the area (in sq. units) of Δ\Delta PTQ is :
A 36536\sqrt 5
B 45545\sqrt 5
C 54354\sqrt 3
D 60360\sqrt 3
Correct Answer
Option B
Solution

Here PQ is the chord of contact. Equation of PQ is x.0 −- y.3 = 36

⇒    y=−12\Rightarrow \,\,\,\,y = - 12

Putting value of y = −- 12 in the equation 4x2 −- y2 = 36 we get , 4x2 −- 144 = 36

⇒    4x2 = 180\Rightarrow \,\,\,\,4{x^2}\, = \,180
⇒     x2  =  45\Rightarrow \,\,\,\,\,{x^2}\,\, = \,\,45
⇒    x=± 35\Rightarrow \,\,\,\,x = \pm \,3\sqrt 5
∴   \therefore\,\,\,

Coordinate of

P=(−35,−12)P = \left( { - 3\sqrt 5 , - 12} \right)

and Coordinate of

Q=(35,−12)Q = \left( {3\sqrt 5 , - 12} \right)
∴   \therefore\,\,\,

Length of PQ

=35+35= 3\sqrt 5 + 3\sqrt 5
=65.= 6\sqrt 5 .

TM is the height of the triangle Length of TM = 12 + 3 = 15

∴   \therefore\,\,\,

Area of

ΔPQT\Delta PQT

=

12×65×15{1 \over 2} \times 6\sqrt 5 \times 15
=455= 45\sqrt 5

sq. units

Q9
Let H\mathrm{H} be the hyperbola, whose foci are (1±2,0)(1 \pm \sqrt{2}, 0) and eccentricity is 2\sqrt{2}. Then the length of its latus rectum is :
A 52\dfrac{5}{2}
B 3
C 2
D 32\dfrac{3}{2}
Correct Answer
Option C
Solution

2ae=∣(1+2)−(1+2)∣=22 2 \mathrm{ae}=|(1+\sqrt{2})-(1+\sqrt{2})|=2 \sqrt{2} ⇒\Rightarrow ae=2\mathrm{ae}=\sqrt{2} ⇒\Rightarrow a=1\mathrm{a}=1 ⇒b=1∵e=2⇒\Rightarrow \mathrm{b}=1 \because \mathrm{e}=\sqrt{2} \Rightarrow Hyperbola is rectangular ⇒L.R=2 b2a=2\Rightarrow \mathrm{L} . \mathrm{R}=\dfrac{2 \mathrm{~b}^{2}}{\mathrm{a}}=2

Q10
If a directrix of a hyperbola centred at the origin and passing through the point (4, –23\sqrt 3 ) is 5x = 45\sqrt 5 and its eccentricity is e, then :
A 4e4 – 24e2 + 27 = 0
B 4e4 – 24e2 + 35 = 0
C 4e4 – 12e2 - 27 = 0
D 4e4 + 8e2 - 35 = 0
Correct Answer
Option B
Solution

5x = 4

5\sqrt 5

⇒\Rightarrow

x=45x = {4 \over {\sqrt 5 }}

Equation of hyperbola

x2a2−y2b2=1{{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1

it passes through

(4,−23)\left( {4, - 2\sqrt 3 } \right)
∴e2=1+b2a2\therefore {e^2} = 1 + {{{b^2}} \over {{a^2}}}
⇒a2e2−a2=b2\Rightarrow {a^2}{e^2} - {a^2} = {b^2}
⇒16a2−12a2e2−a2=1\Rightarrow {{16} \over {{a^2}}} - {{12} \over {{a^2}{e^2} - {a^2}}} = 1
⇒4a2[41−3e2−1]=1\Rightarrow {4 \over {{a^2}}}\left[ {{4 \over 1} - {3 \over {{e^2} - 1}}} \right] = 1
⇒4e2−4−3=(e2−1)(a24)\Rightarrow 4{e^2} - 4 - 3 = ({e^2} - 1)\left( {{{{a^2}} \over 4}} \right)
⇒4(4e3−7)=(e2−1)(4e5)2\Rightarrow 4(4{e^3} - 7) = ({e^2} - 1){\left( {{{4e} \over {\sqrt 5 }}} \right)^2}
⇒4e4−24e2+35=0\Rightarrow 4{e^4} - 24{e^2} + 35 = 0
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