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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Hyperbola

JEE Main · Mathematics · Hyperbola

If the eccentricity of the hyperbola x^ 2 -y^ 2 cosec^ 2 α=25 is 5...

If the eccentricity of the hyperbola x2−y2cosec⁡2α=25x^{2}-y^{2} \operatorname{cosec}^{2} \alpha=25 is 5\sqrt{5} times the eccentricity of the ellipse x2cosec⁡2α+y2=5x^{2} \operatorname{cosec}^{2} \alpha+y^{2}=5 then α\alpha is equal to :

Options

  1. Atan⁡−12\tan ^{-1} \sqrt{2}
  2. Bsin⁡−134\sin ^{-1} \sqrt{\frac{3}{4}}
  3. Ctan⁡−125\tan ^{-1} \sqrt{\frac{2}{5}}
  4. Dsin⁡−125\sin ^{-1} \sqrt{\frac{2}{5}}

Correct answer

A. tan⁡−12\tan ^{-1} \sqrt{2}

Step-by-step solution

Eccentrcity of x225−y225sin⁡2α=1\frac{x^{2}}{25}-\frac{y^{2}}{25 \sin ^{2} \alpha}=1 is 1+sin⁡2α\sqrt{1+\sin ^{2} \alpha}

Eccentricity of x25sin⁡2α+y25=1\frac{x^{2}}{5 \sin ^{2} \alpha}+\frac{y^{2}}{5}=1 \quad is 1−sin⁡2α\sqrt{1-\sin ^{2} \alpha}
Given, 1+sin⁡2α=51−sin⁡2α\sqrt{1+\sin ^{2} \alpha}=\sqrt{5} \sqrt{1-\sin ^{2} \alpha}

<br>⇒sin⁡2α=23<br>⇒α=sin⁡−123=tan⁡−12<br>\begin{array}{l}<br> \Rightarrow \sin ^{2} \alpha=\frac{2}{3}\\<br> \Rightarrow \alpha=\sin ^{-1} \sqrt{\frac{2}{3}}=\tan ^{-1} \sqrt{2}<br> \end{array}
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