Explanation

(b)
Let $P\left( a\ {\cos\theta}_{1,\ }{b\sin\theta}_{1} \right)\ and\ \mathcal{Q}\left( a\ {\cos\theta}_{2,\ }b{\sin\theta}_{2} \right)$ be two point on the ellipse. Then,
$m_{1} = slope\ of\ OP = \frac{b}{a}\tan\theta_{1}\$
and $m_{2} = slope\ of\ O\mathcal{Q} = \frac{b}{a}\ {\tan\theta}_{2}$
[image:11_image7.emf]
$\therefore m_{1}m_{2} = \frac{b^{2}}{a^{2}}{\tan\theta}_{1}{\tan\theta}_{2}$
$= \frac{b^{2}}{a^{2}} \times \frac{{- a}^{2}}{b^{2}}$
$\left\lbrack \therefore{\tan\theta}_{1}\ {\tan\theta}_{2} = - \frac{a^{2}}{b^{2}}(given) \right\rbrack$
$= - 1$
$\therefore < PO\mathcal{Q} = \frac{\pi}{2}$
Hence, $P\mathcal{Q}$ makes a right angle at the centre of the ellipse