The normal at a point $\mathrm{P}\left(\mathrm{\theta }\right)$ on the ellipse $5{\mathrm{x}}^2+14{\mathrm{y}}^2=70$ cuts the again at a point $\mathrm{Q}\left(2\mathrm{\theta }\right)$, then cos$\mathrm{\theta }$=

Normal at ${\mathrm{\prime }}^{\mathrm{\prime }}{\mathrm{\theta }}^{\mathrm{\prime }}\frac{\sqrt{14}\mathrm{x}}{\cos \mathrm{\theta }}-\frac{\sqrt{5}\mathrm{y}}{\sin \mathrm{\theta }}=9$ it passes through $(\sqrt{14}\cos 2\mathrm{\theta },\sqrt{5}\sin 2\mathrm{\theta })$

$\begin{array}{l}Equation of the given ellipse is\\ \frac{x^2}{14}+\frac{y^2}{5}=1\\ Equation of normal is\\ \boxed{\frac{ax}{\cos \theta }-\frac{by}{\sin \theta }=a^2-b^2}\\ \frac{\sqrt{14}x}{\cos \theta }-\frac{\sqrt{5}y}{\sin \theta }=14-5=9\\ \text{ As this passes through }(a\cos 2\theta ,b\sin 2\theta )\\ \begin{array}{l}\therefore \frac{14\cos 2\theta }{\cos \theta }-\frac{5\sin 2\theta }{\sin \theta }=9\\ \therefore 14\left[2{\cos }^2\theta -1\right]-5\times 2\cos \theta =9\cos \theta \\ \therefore 18{\cos }^2\theta -9\cos \theta -14=0\\ \cos \theta =\frac{9\pm \sqrt{81+4\times 14\times 18}}{36}\\ =\frac{9\pm .33}{36}=-\frac{24}{36},\frac{42}{36}\\ \therefore \cos \theta =-\frac{2}{3}\end{array}\end{array}$