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Q.
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
Conic Sections
Solution:
The normal at $P(a \cos \theta, b \sin \theta)$ is
$\frac{a x}{\cos \theta}-\frac{b x}{\sin \theta}=a^{2}-b^{2}$,
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}$