If the tangent at the point $\left(4\cos \mathrm{\varphi },\frac{16}{\sqrt{11}}\sin \mathrm{\varphi }\right)$ the the ellipse $16{\mathrm{x}}^2+11{\mathrm{y}}^2=256$ is also a tangent to the circle ${\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}=15$, then the value of $\mathrm{\varphi }$is 

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

Equation of tangent at P  is $4\cos \mathrm{\varphi x}+\sqrt{11}\sin \mathrm{\varphi y}=16\to \left(1\right)$

$$\begin{gathered} Given circle x^2+y^2-2x-15=0 \\ centre\left(C\right)=(1,0),r=4 \end{gathered}$$
Since (1) is tangent to the circle then r=d

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