Question

If the normal at $\theta$ to the ellipse $5x^2+14y^2=70,$ meets the curve again at $2\theta$, then the value of $|3\cos \theta |$ is

Answer 1

Equation of ellipse : $5x^2+14y^2=70$
$\Rightarrow \frac{x^2}{14}+\frac{y^2}{5}=1$

Equation of normal at $\theta$ is
$\left(a\sec \theta \right)x-\left(b\text{cosec}\theta \right)y=a^2-b^2\Rightarrow \frac{\sqrt{14}x}{\cos \theta }-\frac{\sqrt{5}y}{\sin \theta }=14-5\Rightarrow \frac{\sqrt{14}x}{\cos \theta }-\frac{\sqrt{5}y}{\sin \theta }=9$

Above line also passes through $2\theta$
$Q=(\sqrt{14}\cos 2\theta ,\sqrt{5}\sin 2\theta )$
Putting point $Q$ in the equation of normal, we get
$\frac{\sqrt{14}\left(\sqrt{14}\cos 2\theta \right)}{\cos \theta }-\frac{\sqrt{5}\left(\sqrt{5}\sin 2\theta \right)}{\sin \theta }=9\Rightarrow \frac{14\cos 2\theta }{\cos \theta }-\frac{10\sin \theta \cos \theta }{\sin \theta }=9\Rightarrow 14(2{\cos }^2\theta -1)-10{\cos }^2\theta -9\cos \theta =0\Rightarrow 18{\cos }^2\theta -9\cos \theta -14=0$
$\Rightarrow (3\cos \theta +2)(6\cos \theta -7)=0$
As $\cos \theta \in [-1,1]$, so
$\cos \theta =-\frac{2}{3}\therefore |3\cos \theta |=2$