If the normal at the point  $P\left(\theta \right)$ to the ellipse $5x^2+14y^2=70$ intersect it again at the point  $Q\left(2\theta \right)$, then  $\cos \theta =$

Given ellipse is  $\frac{x^2}{14}+\frac{y^2}{5}=1$
$\therefore P\left(\theta \right)$  is  $(\sqrt{14}\cos \theta ,\sqrt{5}\sin \theta )$
Equation of normal at  $P\left(\theta \right)$ is 
 $\frac{\sqrt{14}x}{\cos \theta }-\frac{\sqrt{5}y}{\sin \theta }=14-5=9$
$Q\left(2\theta \right)$  lies on this normal, therefore 
$$\begin{gathered} \frac{14\cos 2\theta }{\cos \theta }-\frac{5\sin 2\theta }{\sin \theta }=9 \\ \Rightarrow 14\cos 2\theta \sin \theta -10\sin \theta {\cos }^2\theta =9\cos \theta \sin \theta \\ \Rightarrow 18{\cos }^2\theta -9\cos \theta -14=0 \\ \Rightarrow (3\cos \theta +2)(6\cos \theta -7)=0 \\ \Rightarrow \cos \theta =-\frac{2}{3}or\frac{7}{6} \\ \Rightarrow \cos \theta =-\frac{2}{3}as\cos \theta \ne \frac{7}{6} \end{gathered}$$