If the tangent at the point $\left(4 \cos 2 \theta, \frac{16}{\sqrt{11}} \sin 2 \theta\right)$ on the…
If the tangent at the point $\left(4 \cos 2 \theta, \frac{16}{\sqrt{11}} \sin 2 \theta\right)$ on the ellipse $16 x^2+11 y^2=256$ touches the circle $x^2+y^2-2 x=15$, then $\theta=$
$\pm \frac{\pi}{3}$
$\pm \frac{\pi}{6}$
$\pm \frac{\pi}{4}$
$\pm \frac{\pi}{8}$
Solution
Equation of tangent at point $\left(4 \cos 2 \theta, \frac{16}{\sqrt{11}} \sin 2 \theta\right)$ on the ellipse $16 x^2+11 y^2=256$ is
$
\begin{aligned}
& 16 x(4 \cos 2 \theta)+11 y\left(\frac{16}{\sqrt{11}} \sin 2 \theta\right)=256 \\
& \Rightarrow \quad 4 x \cos 2 \theta+\sqrt{11} y \sin 2 \theta=16
\end{aligned}
$
Since, tangent (i) is tangent to the circle
$
\begin{array}{ll}
& x^2+y^2-2 x-15=0 \text {, so } \\
& \frac{|4 \cos 2 \theta-16|}{\sqrt{16 \cos ^2 2 \theta+11 \sin ^2 2 \theta}}=4 \\
\Rightarrow \quad & \cos ^2 2 \theta-8 \cos 2 \theta+16=11+5 \cos ^2 2 \theta \\
\Rightarrow & 4 \cos ^2 2 \theta+8 \cos 2 \theta-5=0 . \\
\Rightarrow & \cos 2 \theta=\frac{1}{2}=\cos \frac{\pi}{3} \text { or } \cos \left(-\frac{\pi}{3}\right) \\
\Rightarrow & \quad \theta= \pm \frac{\pi}{6}
\end{array}
$