If the tangent at $P\left(\theta \right)$ on the ellipse $16x^2+11y^2=256$

touches the circle  $x^2+y^2+2x-15=0,$ then $\theta -$

Centre of the circle is (– 1, 0) and radius is 4. Equation of the tangent at $P\left(\theta \right)$ to the ellipse is

$\frac{x}{4}\cos \theta +\frac{y}{16/\sqrt{11}}\sin \theta =1$

If this touches the circle, then

$\begin{array}{l}\frac{\frac{-\cos \theta }{4}-1}{\sqrt{\frac{{\cos }^2 \theta }{16}+\frac{{\sin }^2 \theta }{256/11}}}=4\\ \Rightarrow (\cos \theta +4{)}^2=256\left(\frac{{\cos }^2 \theta }{16}+\frac{11{\sin }^2 \theta }{256}\right)\\ \Rightarrow 4{\cos }^2 \theta -8\cos \theta -5=0\\ \Rightarrow (2\cos \theta -5)(2\cos \theta +1)=0\\ \Rightarrow \cos \theta =-1/2\Rightarrow \theta =2\pi /3\end{array}$