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If the normal at the point $P(\theta )$ to the ellipse $\frac{{{x^2}}}{{14}} + \frac{{{y^2}}}{5} = 1$ intersects it again at the point $Q(2\theta )$, then $\cos \theta $ is equal to
$\frac{2}{3}$
$ - \frac{2}{3}$
$\frac{3}{2}$
$ - \frac{3}{2}$
Solution
(b) The normal at $P(a\cos \theta ,b\sin \theta )$ is $ax\sec \theta – by\,\,{\rm{cosec}}\theta = {a^2} – {b^2}$, 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 18{\cos ^2}\theta – 9\cos \theta – 14 = 0$
$ \Rightarrow (6\cos \theta – 7)(3\cos \theta + 2) = 0 $
$\Rightarrow \cos \theta = – \frac{2}{3}$.
