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

$$\begin{gathered} \mathrm{Equation} \mathrm{of} \mathrm{tangent} \mathrm{is} \\ \frac{\mathrm{xcos}\mathrm{\varphi }}{4}+\frac{\sqrt{11}\sin \mathrm{\varphi }}{16}\mathrm{y}=1 \\ \Rightarrow 4x\cos \mathrm{\varphi }+\sqrt{11}y\sin \mathrm{\varphi }=16 \end{gathered}$$
Is also tangent to $(\mathrm{x}-1{)}^2+{\mathrm{y}}^2=16$

So the perpendicular distance from centre i.e (1,0) to the tangent is equal to the radius of the circle

$\begin{array}{r}\frac{\left|4\cos \varphi -16\right|}{\sqrt{16{\cos }^2\varphi +11{\sin }^2\varphi }}=4\\ \Rightarrow {\left(\cos \varphi -4\right)}^2=16{\cos }^2\varphi +11{\sin }^2\varphi \\ \Rightarrow {\cos }^2\varphi -8\cos \varphi +16=5{\cos }^2\varphi +11\\ \Rightarrow 4{\cos }^2\varphi +8\cos \varphi -5=0\\ \Rightarrow \cos \varphi =\frac{1}{2},\frac{-5}{2}\\ \Rightarrow \cos \varphi =\frac{-5}{2}\text{ is not possible }\\ \Rightarrow \cos \varphi =\frac{1}{2}\\ \Rightarrow \varphi =\frac{\pi }{3},\frac{5\pi }{3}\end{array}$