Question

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

• A. $\frac{2}{3}$
• B. $-\frac{2}{3}$
• C. $\frac{3}{2}$
• D. $-\frac{3}{2}$

Answer 1

Answer: (A)

The normal at $P(a\cos \theta ,b\sin \theta )$ is
$\frac{ax}{\cos \theta }-\frac{bx}{\sin \theta }=a^2-b^2$,
where $a^2=14,b^2=5$
It meets the curve again at $Q(2\theta )$,
i.e., $(a\cos 2\theta ,b\sin 2\theta )$
$\therefore \frac{a}{\cos \theta }(a\cos 2\theta )-\frac{b}{\sin \theta }(b\sin 2\theta )=a^2-b^2$
$\Rightarrow \frac{14}{\cos \theta }(\cos 2\theta )-\frac{5}{\sin \theta }(\sin 2\theta )=14-5$
$\Rightarrow 28{\cos }^2\theta -14-10{\cos }^2\theta =9\cos \theta$
$\Rightarrow 18{\cos }^2\theta -9\cos \theta -14=0$
$\Rightarrow (6\cos \theta -7)(3\cos \theta -2)=0$
$\Rightarrow \cos \theta =\frac{2}{3}$