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

Given ellipse $\mathrm{S}\equiv 16{\mathrm{x}}^2+11{\mathrm{y}}^2-256=0$

Given point $\mathrm{P}=\left(4\cos 2\theta , \frac{16}{\sqrt{11}} \sin 2\mathrm{\theta }\right)$

Equation of tangent at $\mathrm{P}\left(\mathrm{\theta }\right)$ is ${\mathrm{S}}_1=0$

$$\begin{gathered} \Rightarrow 16x\left(4\cos 2\mathrm{\theta }\right)+11\left(\frac{16}{\sqrt{11}}\right) \left(\sin 2\mathrm{\theta }\right)y-256=0 \\ \Rightarrow 4\mathrm{x} \cos 2\mathrm{\theta }+\left(\sqrt{11} \sin 2\mathrm{\theta }\right) \mathrm{y}-16=0 \end{gathered}$$

since it touches the circle ${\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-15=0$ then r=d where $C=(1, 0), \mathrm{r}=\sqrt{16}$

$$\begin{gathered} \Rightarrow \sqrt{16}=\frac{|4\cos 2\mathrm{\theta }-16|}{\sqrt{16{\cos }^{2}2\mathrm{\theta }+11{\sin }^{2}2\mathrm{\theta }}} \\ \Rightarrow 16(16{\cos }^{2}2\mathrm{\theta }+11{\sin }^{2}2\mathrm{\theta })=16{(\cos 2\mathrm{\theta }-4)}^2 \\ \Rightarrow 16{\cos }^{2}2\mathrm{\theta }+11-11{\cos }^{2}2\mathrm{\theta }={\cos }^{2}2\mathrm{\theta }+16-8\cos 2\mathrm{\theta } \\ \Rightarrow 4{\cos }^{2}2\mathrm{\theta }+8\cos 2\mathrm{\theta }-5=0 \\ \Rightarrow 4{\cos }^{2}2\mathrm{\theta }+10\cos 2\mathrm{\theta }-2\cos 2\mathrm{\theta }-5=0 \\ \Rightarrow (2\mathrm{cos\theta }+5) (2\mathrm{cos\theta }-1)=0 \\ \Rightarrow \mathrm{cos\theta }=\frac{-5}{2} \left(\mathrm{or}\right) \mathrm{cos\theta }=\frac{1}{2} \\ \Rightarrow \mathrm{cos\theta }=\frac{1}{2} (\because -1\le \mathrm{cos\theta }\le 1) \\ \Rightarrow \theta =n\pi \pm \frac{\mathrm{\pi }}{3},n\in Z \\ If n=0 then \mathrm{\theta }=\pm \frac{\mathrm{\pi }}{3} \end{gathered}$$