Question

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

• A. $\pm \frac{\pi }{4}$
• B. $\pm \frac{\pi }{3}$
• C. $\pm \frac{\pi }{6}$
• D. $\pm \frac{\pi }{2}$

Answer 1

Answer: (B)

The equation of the tangent at $\left(4\cos \theta ,\frac{16}{\sqrt{11}}\sin \theta \right)$ to
the ellipse $16x^2+11y^2=256$ is
$16x(4\cos \theta )+11y\left(\frac{16}{\sqrt{11}}\sin \theta \right)=256$
or $4x\cos \theta +\sqrt{11}y\sin \theta =16$
This touches the circle $(x-1{)}^2+y^2=4^2$ if
$\left|\frac{4\cos \theta -16}{\sqrt{16{\cos }^2\theta +11{\sin }^2\theta }}\right|=4$
$\Rightarrow (\cos \theta -4{)}^2=16{\cos }^2\theta +11{\sin }^2\theta$
$\Rightarrow 15{\cos }^2\theta +11{\sin }^2\theta +8\cos \theta -16=0$
$\Rightarrow 4{\cos }^2\theta +8\cos \theta -5=0$
$\Rightarrow (2\cos \theta -1)(2\cos \theta +5)=0$
$\Rightarrow \cos \theta =\frac{1}{2}$
$\Rightarrow \theta =\pm \frac{\pi }{3}$