Question

$P$ is point on ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Its polar line w.r.t. hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ meets the asymptotes of the hyperbola in $Q$ and $R$. If $QR=c$, then the value of eccentric angle $\theta$ of $P$ is given by:

• A. $c^2{\cos }^{2}2\theta +2(a^2-b^2){\cos }^{2}2\theta =2(a^2+b^2)$
• B. $c^2{\cos }^2\theta +2(a^2-b^2)\cos \theta =2(a^2+b^2)$
• C. $c^2\cos 2\theta +2(a^2-b^2)\cos 2\theta =2(a^2+b^2)$
• D. $c^2{\cos }^{2}2\theta +2(a^2-b^2)\cos 2\theta =2(a^2+b^2)$

Answer 1

Answer: (D)

$c^2{\cos }^{2}2\theta +2(a^2-b^2)\cos 2\theta =2(a^2+b^2)$

Let the point $P$ be $(a\cos \theta ,b\sin \theta )$

So, the polar line w.r.t hyperbola will be:

$\frac{xa\cos \theta }{a^2}-\frac{yb\sin \theta }{b^2}=1$

Or, $\frac{x\cos \theta }{a}-\frac{y\sin \theta }{b}=1$

The asymptotes of the hyperbola is:

$y=\pm \frac{bx}{a}$

The point of intersection $Q$ will be:

$\frac{x\cos \theta }{a}-\frac{bx\sin \theta }{ab}=1$

Or, $x=\frac{a}{\cos \theta -\sin \theta }$ and $y=\frac{b}{\cos \theta -\sin \theta }$

And, the point of intersection $R$ will be:

$\frac{x\cos \theta }{a}+\frac{bx\sin \theta }{ab}=1$

Or, $x=\frac{a}{\cos \theta +\sin \theta }$ and $y=\frac{-b}{\cos \theta +\sin \theta }$

Given, $QR=c$ Or $\frac{a^2{(\cos \theta +\sin \theta -\cos \theta +\sin \theta )}^2}{((\cos \theta {)}^2-{\left(\sin \theta \right)}^2)}+\frac{b^2{(\cos \theta +\sin \theta +\cos \theta -\sin \theta )}^2}{((\cos \theta {)}^2-{\left(\sin \theta \right)}^2)}=c^2$

Or $4a^2{\left(\sin \theta \right)}^2+{4b}^2(\cos \theta {)}^2=c^2(\cos 2\theta {)}^2$

Or ${4b}^2+4(a^2-b^2){\left(\sin \theta \right)}^2=c^2(\cos 2\theta {)}^2$

Or ${-4b}^2-4(a^2-b^2){\left(\sin \theta \right)}^2={-c}^2(\cos 2\theta {)}^2$

Adding and subtracting $2(a^2-b^2)$, we get

${-4b}^2-2(a^2-b^2)+2(a^2-b^2)(1-2{\left(\sin \theta \right)}^2)={-c}^2(\cos 2\theta {)}^2$

$2(-a^2-b^2)+2(a^2-b^2)(\cos 2\theta )={-c}^2(\cos 2\theta {)}^2$

$c^2(\cos 2\theta {)}^2+2(a^2-b^2)(\cos 2\theta )=2(a^2+b^2)$