Question

If the normal at the point $P\left(\theta \right)$ to the ellipse $\frac{x^2}{14}+\frac{y^2}{5}=1$ intersects it again at the point $Q\left(2\theta \right)$, then

• A. $\sin \theta =-\frac{2}{3}$
• B. $\sin \theta =\frac{3}{4}$
• C. $\cos \theta =\frac{3}{4}$
• D. $\cos \theta =-\frac{2}{3}$

Answer 1

The correct option is D $\cos \theta =-\frac{2}{3}$

Equation of ellipse is
$\frac{x^2}{14}+\frac{y^2}{5}=1$
$P\left(\theta \right)\equiv (\sqrt{14}\cos \theta ,\sqrt{5}\sin \theta )$

The normal at $P$ will be
$\frac{\sqrt{14}x}{\cos \theta }-\frac{\sqrt{5}x}{\sin \theta }=14-5\Rightarrow \frac{\sqrt{14}x}{\cos \theta }-\frac{\sqrt{5}x}{\sin \theta }=9$

If it meets the curve again at $Q\left(2\theta \right)\equiv (\sqrt{14}\cos 2\theta ,\sqrt{5}\sin 2\theta ),$ then

$\frac{\sqrt{14}\left(\sqrt{14}\cos 2\theta \right)}{\cos \theta }-\frac{\sqrt{5}\left(\sqrt{5}\sin 2\theta \right)}{\sin \theta }=9$
$\Rightarrow \frac{14(2{\cos }^2\theta -1)}{\cos \theta }-5\left(2\cos \theta \right)=9$
$\Rightarrow 28{\cos }^2\theta -14-10{\cos }^2\theta =9\cos \theta$
$\Rightarrow 18{\cos }^2\theta -9\cos \theta -14=0$
$\Rightarrow (6\cos \theta -7)(3\cos \theta +2)=0$
$\Rightarrow \cos \theta =-\frac{2}{3}\text{or}\cos \theta =\frac{7}{6}\left(\text{Not possible}\right)$
Hence, $\cos \theta =-\frac{2}{3}$