Question

If the normal at $\theta$ on the ellipse $5x^2+14y^2=70$ cuts the curve again at a point $2\theta$, then $\cos \theta$ =

• A. $\frac{2}{3}$
• B. $-\frac{2}{3}$
• C. $\frac{1}{3}$
• D. $-\frac{1}{3}$

Answer 1

The correct option is B $-\frac{2}{3}$

Equation of ellipse is $\frac{x^2}{14}+\frac{y^2}{5}=1$
Equation of normal at $\theta$ is $\frac{\sqrt{14}x}{\cos \theta }-\frac{\sqrt{5}y}{\sin \theta }=9$
Since it passes through $(\sqrt{14}\cos 2\theta ,\sqrt{5}\sin 2\theta )$ then $\frac{14.\cos 2\theta }{\cos \theta }-\frac{5.\sin 2\theta }{\sin \theta }=9$
$\Rightarrow 14.(2{\cos }^2\theta -1)-10{\cos }^2\theta =9\cos \theta \Rightarrow 18{\cos }^2\theta -9\cos \theta -14=0\Rightarrow \cos \theta =-\frac{2}{3}$