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JEE Advanced 2007Paper 1
Exam
JEE Advanced
Year
2007
Paper
Paper 1
Subject
Mathematics
Chapter
Ellipse

JEE Advanced 2007 · Paper 1 · Mathematics · Ellipse

A hyperbola, having the transverse axis of length 2 θ , is confocal...

A hyperbola, having the transverse axis of length 2sin⁡θ2 \sin \theta, is confocal with the ellipse 3x2+4y2=123 x^2+4 y^2=12. Then, its equation is

Options

  1. Ax2cosec⁡2θ−y2sec⁡2θ=1x^2 \operatorname{cosec}^2 \theta-y^2 \sec ^2 \theta=1
  2. Bx2sec⁡2θ−y2cosec⁡2θ=1x^2 \sec ^2 \theta-y^2 \operatorname{cosec}^2 \theta=1
  3. Cx2sin⁡2θ−y2cos⁡2θ=1x^2 \sin ^2 \theta-y^2 \cos ^2 \theta=1
  4. Dx2cos⁡2θ−y2sin⁡2θ=1x^2 \cos ^2 \theta-y^2 \sin ^2 \theta=1

Correct answer

A. x2cosec⁡2θ−y2sec⁡2θ=1x^2 \operatorname{cosec}^2 \theta-y^2 \sec ^2 \theta=1

Step-by-step solution

The given ellipse is x24+y23=1⇒α=2,β=3∴3=4(1−e2)⇒e=12\begin{aligned}& \frac{x^2}{4}+\frac{y^2}{3}=1 \Rightarrow \alpha=2, \beta=\sqrt{3}\\ & \therefore \quad 3=4 \left(1-e²\right) \Rightarrow e=\frac{1}{2}\\ & \end{aligned}\therefore \quad a e=1 Hencetheeccentricitye1the eccentricitye_1, of the hyperbola is given by 1=e1sin⁡θ⇒e1=cosec⁡θ⇒b2=sin⁡2θ(cosec⁡2θ−1)=cos⁡2θ\begin{array}{rlrl}& & 1 & =e_1 \sin \theta\\ \Rightarrow & & e_1 & =\operatorname{cosec} \theta\\ \Rightarrow & b² & =\sin ^2 \theta \left(\operatorname{cosec}^2 \theta-1\right)=\cos ^2 \theta \end{array}Hence,thehyperbolaisx2sin⁡2θthe hyperbola is\frac{x^2}{\sin ^2 \theta}-y2cos⁡2θ=\frac{y^2}{\cos ^2 \theta}=1  or \text{ or }x^2 \operatorname{cosec}^2θ−\theta-y^2 \sec ^2 θ=\theta=1
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