Search any question & find its solution
Question:
Answered & Verified by Expert
$P Q$ is a double ordinate of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ such that $\triangle O P Q$ is an equilateral triangle, $O$ being the centre of the hyperbola. Then the eccentricity e of the hyperbola satisfies
Options:
Solution:
2182 Upvotes
Verified Answer
The correct answer is:
$\mathrm{e} > 2 / \sqrt{3}$
$\tan 30^{\circ}=\frac{b \tan \theta}{a \sec \theta}$
$\Rightarrow \frac{b}{a}=\frac{1}{\sin \theta \sqrt{3}}$
$e=\sqrt{1+\frac{1}{3 \sin ^2 \theta}} > \sqrt{1+\frac{1}{3}}$
$\Rightarrow e > \frac{2}{\sqrt{3}}\left(0 < \sin ^2 \theta < 1\right)$
$\Rightarrow \frac{b}{a}=\frac{1}{\sin \theta \sqrt{3}}$
$e=\sqrt{1+\frac{1}{3 \sin ^2 \theta}} > \sqrt{1+\frac{1}{3}}$
$\Rightarrow e > \frac{2}{\sqrt{3}}\left(0 < \sin ^2 \theta < 1\right)$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.