Search any question & find its solution
Question:
Answered & Verified by Expert
If $P Q$ is a double ordinate of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ such that $\Delta O P Q$ is equilateral. $O$ being the centre. Then, the eccentricity $e$ satisfies
Options:
Solution:
1903 Upvotes
Verified Answer
The correct answer is:
$e>\frac{2}{\sqrt{3}}$
Given equation of hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ and $\Delta O P Q$ is an equlateral triangle. $P Q$ is double ordinate of the hyperbola.
Let the coordinates of $P$ and $Q$ be $(a \sec \theta, b$ tan $\theta)$ and $(a \sec \theta,-b$ tane $),$ respectively.
In$\Delta O P Q$,
$O P=P Q$
$\Rightarrow \quad a^{2} \sec ^{2} \theta+b^{2} \tan ^{2} \theta=(2 b \tan \theta)^{2}$
$\Rightarrow \quad a^{2} \sec ^{2} \theta=3 b^{2} \tan ^{2} \theta$
$\Rightarrow \quad \sin ^{2} \theta=\frac{a^{2}}{3 b^{2}}$
Now, $\sin ^{2} \theta < 1$
$\Rightarrow \quad \frac{a^{2}}{3 b^{2}} < 1$
$\frac{b^{2}}{a^{2}}>\frac{1}{3}$
$\Rightarrow \quad 1+\frac{b^{2}}{a^{2}}>\frac{4}{3}$
$\therefore$
$e^{2}>\frac{4}{3} \Rightarrow e>\frac{2}{\sqrt{3}}$
Let the coordinates of $P$ and $Q$ be $(a \sec \theta, b$ tan $\theta)$ and $(a \sec \theta,-b$ tane $),$ respectively.
In$\Delta O P Q$,
$O P=P Q$
$\Rightarrow \quad a^{2} \sec ^{2} \theta+b^{2} \tan ^{2} \theta=(2 b \tan \theta)^{2}$
$\Rightarrow \quad a^{2} \sec ^{2} \theta=3 b^{2} \tan ^{2} \theta$
$\Rightarrow \quad \sin ^{2} \theta=\frac{a^{2}}{3 b^{2}}$
Now, $\sin ^{2} \theta < 1$
$\Rightarrow \quad \frac{a^{2}}{3 b^{2}} < 1$
$\frac{b^{2}}{a^{2}}>\frac{1}{3}$
$\Rightarrow \quad 1+\frac{b^{2}}{a^{2}}>\frac{4}{3}$
$\therefore$
$e^{2}>\frac{4}{3} \Rightarrow e>\frac{2}{\sqrt{3}}$
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.