If the tangents at the point $\left(4\cos 2\mathrm{\theta },\frac{16}{\sqrt{11}}\sin 2\mathrm{\theta }\right)$ on the ellipse $16{\mathrm{x}}^2+17{\mathrm{y}}^2=256$ touches the circle ${\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}=15$ then $\mathrm{\theta }=$

$$\begin{gathered} Given ellipse \frac{x^2}{16}+\frac{y^2}{{\displaystyle \frac{256}{11}}}=1 \\ Given point P=(4\cos 2\theta ,\frac{16}{\sqrt{11}}\sin 2\theta ) \end{gathered}$$

equation  of Tangent is $\frac{\mathrm{x}}{4} \cos 2\mathrm{\theta }+\frac{\mathrm{y}}{{\displaystyle \frac{16}{\sqrt{11}}}} \sin 2\mathrm{\theta }-1=0$ it touches  the cirlcle ${\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-15=0 with \mathrm{centre}(1, 0)$

and radius$=\sqrt{1+15}=4$ then d=r

$\perp$ distance from centre to Tangent is equal to raduis

$$\begin{gathered} \left|\frac{4 \cos 2\mathrm{\theta }-16}{\sqrt{16 {\cos }^{2}2\mathrm{\theta }+11 {\sin }^{2}2\mathrm{\theta }}}\right|=4 \\ \left|\cos 2\mathrm{\theta }-4\right|=\sqrt{11+5 {\cos }^{2}2\mathrm{\theta }} \\ \mathrm{S}.\mathrm{B}.\mathrm{S} \\ {\cos }^{2}2\mathrm{\theta }+16-8 \cos 2\mathrm{\theta } \\ =11+5 {\cos }^{2}2\mathrm{\theta } \\ 4 {\cos }^{2}2\mathrm{\theta }+8\cos 2\mathrm{\theta }-5=0 \\ 4 {\cos }^{2}2\mathrm{\theta }+10 \cos 2\mathrm{\theta }-2\cos 2\mathrm{\theta }-5=0 \\ (2 \cos 2\mathrm{\theta }-1) (2 \cos 2\mathrm{\theta }+5)=0 \\ \cos 2\mathrm{\theta }=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ 2\mathrm{\theta }=\pm \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$3$}\right. \Rightarrow \mathrm{\theta }=\pm \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$6$}\right. \end{gathered}$$