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 $\theta$ =

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

Answer 1

The correct option is D $\frac{\pi }{3}$

Given ellipse equation is $\frac{x^2}{16}+\frac{11y^2}{16}=1$

The equation of tangent at $P\left(\theta \right)$ is $\frac{x\cos \theta }{4}+\frac{\sqrt{11}y\sin \theta }{4}=1$

This tangent is also a tangent to the circle $(x-1{)}^2+y^2=16$

$⟹\frac{\frac{\cos \theta }{4}-1}{\sqrt{\frac{{\cos }^2\theta }{16}+\frac{11{\sin }^2\theta }{16}}}=4$

$(\frac{\cos \theta -4}{4}{)}^2=16(\frac{{\cos }^2\theta }{16}+\frac{11{\sin }^2\theta }{16})$

$⟹\cos \theta =\frac{1}{2}⟹\theta =\frac{\pi }{3}$