If the tangent at Point P to the ellipse $16{\mathrm{x}}^2+11{\mathrm{y}}^2=256$ is also the tangent to the circle ${\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}=15$, then the eccentric angle of point P is

$$\begin{gathered} Given ellipse \frac{x^2}{16}+\frac{y^2}{{\displaystyle \frac{256}{11}}}=1 \\ where a^2=16,b^2=\frac{256}{11} \\ Let P=(4\cos \theta ,\frac{16}{\sqrt{11}}\sin \theta ) be any point on given ellipse \end{gathered}$$

Equation of tangent at P is 
$\frac{\mathrm{x}}{4}\cos \mathrm{\theta }+\frac{\sqrt{11}}{16}\sin \mathrm{\theta }\cdot \mathrm{y}=1 ...\left(2\right)$
It is a tangent of circle  ${\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-15=0$ then r = d

where centre=(1,0) and r=4

$$\begin{gathered} \Rightarrow 4=\frac{\left|{\displaystyle \frac{\cos \theta }{4}}-1\right|}{\sqrt{{\displaystyle \frac{{\cos }^2\theta }{16}}+{\displaystyle \frac{11}{256}}{\sin }^2\theta }} \\ squaring on both sides \\ 16(\frac{{\cos }^2\theta }{16}+\frac{11}{256}{\sin }^2\theta )=\frac{{(\cos \theta -4)}^2}{16} \\ \Rightarrow 16{\cos }^2\theta +11{\sin }^2\theta ={\cos }^2\theta +16-8\cos \theta \\ \Rightarrow 4{\cos }^2\theta +8\cos \theta -5=0 \\ \Rightarrow (2\cos \theta +5)(2\cos \theta -1)=0 \\ \Rightarrow \cos \theta =\frac{1}{2}(\because -1\le \cos \theta \le 1) \end{gathered}$$

$$\begin{gathered} \Rightarrow \theta =2n\pi \pm \frac{\pi }{3},n\in z \\ put n=0 then \theta =\pm \frac{\pi }{3} \end{gathered}$$