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Hyperbola question

2019 · 9 Jan · Shift 1 · Q29
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  5. /2019 · 9 Jan · Shift 1 · Q29

Hyperbola question

2019 · 9 Jan · Shift 1 · Q29

JEE MainMathematicsHyperbolaMCQ+4 / −1
Let 0<θ<π20 \lt \theta \lt {\pi \over 2}0<θ<2π​. If the eccentricity of the hyperbola x2cos⁡2θ−y2sin⁡2θ{{{x^2}} \over {{{\cos }^2}\theta }} - {{{y^2}} \over {{{\sin }^2}\theta }}cos2θx2​−sin2θy2​ = 1 is greater than 2, then the length of its latus rectum lies in the interval :
  1. A
    (3, ∞\infty∞)
  2. B
    (32,2]\left( {{3 \over 2},2} \right](23​,2]
  3. C
    (1,32]\left( {1,{3 \over 2}} \right](1,23​]
  4. D
    (2,3]\left( {2,3} \right](2,3]
View written solutionFree

Correct answer: A

  1. Identify the hyperbola parameters

Given

x2cos⁡2θ−y2sin⁡2θ=1,0<θ<π2\frac{x^2}{\cos^2\theta}-\frac{y^2}{\sin^2\theta}=1, \qquad 0<\theta<\frac\pi2cos2θx2​−sin2θy2​=1,0<θ<2π​

This is of the standard form

x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1a2x2​−b2y2​=1

with

a2=cos⁡2θ,b2=sin⁡2θ.a^2=\cos^2\theta,\qquad b^2=\sin^2\theta.a2=cos2θ,b2=sin2θ.

Since 0<θ<π20<\theta<\frac\pi20<θ<2π​, both sin⁡θ\sin\thetasinθ and cos⁡θ\cos\thetacosθ are positive.

  1. Find the eccentricity

For the hyperbola

x2a2−y2b2=1,\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,a2x2​−b2y2​=1,

we have

c2=a2+b2,c^2=a^2+b^2,c2=a2+b2,

and eccentricity

e=ca=1+b2a2.e=\frac ca=\sqrt{1+\frac{b^2}{a^2}}.e=ac​=1+a2b2​​.

So here,

e=1+sin⁡2θcos⁡2θ=1+tan⁡2θ=sec⁡θe=\sqrt{1+\frac{\sin^2\theta}{\cos^2\theta}} =\sqrt{1+\tan^2\theta} =\sec\thetae=1+cos2θsin2θ​​=1+tan2θ​=secθ

because θ∈(0,π/2)\theta\in(0,\pi/2)θ∈(0,π/2).

Given that eccentricity is greater than 222,

sec⁡θ>2  ⟹  cos⁡θ<12.\sec\theta>2 \implies \cos\theta<\frac12.secθ>2⟹cosθ<21​.

Since 0<θ<π20<\theta<\frac\pi20<θ<2π​,

0<cos⁡θ<12.0<\cos\theta<\frac12.0<cosθ<21​.

Hence

cos⁡2θ<14.\cos^2\theta<\frac14.cos2θ<41​.

Then

sin⁡2θ=1−cos⁡2θ>1−14=34.\sin^2\theta=1-\cos^2\theta>1-\frac14=\frac34.sin2θ=1−cos2θ>1−41​=43​.
  1. Length of latus rectum

For the hyperbola

x2a2−y2b2=1,\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,a2x2​−b2y2​=1,

the length of the latus rectum is

2b2a.\frac{2b^2}{a}.a2b2​.

Here,

a=cos⁡θ,b2=sin⁡2θ.a=\cos\theta,\qquad b^2=\sin^2\theta.a=cosθ,b2=sin2θ.

Therefore,

L=2sin⁡2θcos⁡θ.L=\frac{2\sin^2\theta}{\cos\theta}.L=cosθ2sin2θ​.

Using sin⁡2θ=1−cos⁡2θ\sin^2\theta=1-\cos^2\thetasin2θ=1−cos2θ,

L=2(1−cos⁡2θ)cos⁡θ=2(1cos⁡θ−cos⁡θ).L=\frac{2(1-\cos^2\theta)}{\cos\theta} =2\left(\frac1{\cos\theta}-\cos\theta\right).L=cosθ2(1−cos2θ)​=2(cosθ1​−cosθ).

Let

x=cos⁡θ,0<x<12.x=\cos\theta,\qquad 0<x<\frac12.x=cosθ,0<x<21​.

Then

L=2(1x−x).L=2\left(\frac1x-x\right).L=2(x1​−x).
  1. Find the interval of LLL

Consider

f(x)=2(1x−x),0<x<12.f(x)=2\left(\frac1x-x\right), \qquad 0<x<\frac12.f(x)=2(x1​−x),0<x<21​.

Its derivative is

f′(x)=2(−1x2−1)<0.f'(x)=2\left(-\frac1{x^2}-1\right)<0.f′(x)=2(−x21​−1)<0.

So f(x)f(x)f(x) is strictly decreasing on (0,1/2)(0,1/2)(0,1/2).

  • As x→12−x\to \frac12^{-}x→21​−,
L→2(2−12)=2⋅32=3.L\to 2\left(2-\frac12\right)=2\cdot\frac32=3.L→2(2−21​)=2⋅23​=3.

But x=12x=\frac12x=21​ is not allowed because e>2e>2e>2 (strict inequality), so L=3L=3L=3 is not included.

  • As x→0+x\to 0^+x→0+,
L→∞.L\to \infty.L→∞.

Therefore,

L∈(3,∞).L\in (3,\infty).L∈(3,∞).
  1. Check options
  • A: (3,∞)(3,\infty)(3,∞) ✅
  • B: (32,2]\left(\frac32,2\right](23​,2] ❌
  • C: (1,32]\left(1,\frac32\right](1,23​] ❌
  • D: (2,3](2,3](2,3] ❌

Hence the correct option is

A\boxed{A}A​
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