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

$a^2=14,b^2=5$

The normal at $\theta$ is $\frac{ax}{\cos \theta }-\frac{by}{\sin \theta }=a^2-b^2$

It passes through the point $2\theta$

$\therefore \frac{14\cos 2\theta }{\cos \theta }-\frac{5\sin 2\theta }{\sin \theta }=9\Rightarrow 18{\cos }^2\theta -9\cos \theta -14=0$

solving, $\cos \theta =-\frac{2}{3}\text{.}$ The other root  $\cos \theta =\frac{7}{6}$ is not valid,