Question

If the tangent at the point $P\left(\theta \right)$ to the ellipse $16x^2+11y^2=256$ is also a tangent to the circle $x^2+y^2-2x=15,$ then possible value(s) of $\theta$ is/are

• A. $\frac{2\pi }{3}$
• B. $\frac{4\pi }{3}$
• C. $\frac{5\pi }{3}$
• D. $\frac{\pi }{3}$

Answer 1

Answer: (C), (D)

$\frac{\pi }{3}$

Given circle is $x^2+y^2-2x=15$
$\Rightarrow x^2-2x+1+y^2=15+1\Rightarrow (x-1{)}^2+y^2=4^2$

Ellipse is $16x^2+11y^2=256$
For ellipse, equation of tangent at $(4\cos \theta ,\frac{16}{\sqrt{11}}\sin \theta )$ is:
$16x\left(4\cos \theta \right)+11y\left(\frac{16}{\sqrt{11}}\sin \theta \right)=256\Rightarrow x\left(4\cos \theta \right)+11y\left(\frac{1}{\sqrt{11}}\sin \theta \right)=16$

It touches the circle $(x-1{)}^2+y^2=4^2$
So, distance of tangent from centre of circle is $4$
$\Rightarrow \left|\frac{4\cos \theta -16}{\sqrt{16{\cos }^2\theta +11{\sin }^2\theta }}\right|=4\Rightarrow (\cos \theta -4{)}^2=16{\cos }^2\theta +11{\sin }^2\theta \Rightarrow 4{\cos }^2\theta +8\cos \theta -5=0\Rightarrow \cos \theta =\frac{1}{2}\text{or}\cos \theta =-\frac{5}{2}\text{(not possible)}$
$\therefore \theta =2n\pi \pm \frac{\pi }{3},n\in Z$

Hence, possible values of $\theta$ are $\frac{\pi }{3},\frac{5\pi }{3}$