Question

The point(s) on the ellipse $16x^2+11y^2=256$
where the common tangent to it and the circle $x^2+y^2-2x=15$ touch is

• A. $(2,8\sqrt{\frac{3}{11}})$
• B. $(2,8\sqrt{\frac{3}{11}})$
• C. $(-2,-8\sqrt{\frac{3}{11}})$
• D. $(2,-8\sqrt{\frac{3}{11}})$

Answer 1

Answer: (A), (D)

The correct options are
A $(2,8\sqrt{\frac{3}{11}})$
D $(2,-8\sqrt{\frac{3}{11}})$

Ellipse:
$\frac{x^2}{16}+\frac{11y^2}{{16}^2}=1$
Tangent to the ellipse at point
$P(4\cos \theta ,\frac{16}{\sqrt{11}}\sin \theta )$ is
$\frac{x}{4}\cos \theta +\frac{\sqrt{11}y}{16}\sin \theta =1$
It also touches the circle $x^2+y^2-2x=15(x-1{)}^2+y^2=4^2$

$\therefore$distance of the tangent to the centre of the circle = radius
$\therefore \frac{|\frac{1}{4}\cos \theta -1|}{\sqrt{\frac{{\cos }^2\theta }{16}+\frac{11{\sin }^2\theta }{256}}}=4|\cos \theta -4|=\sqrt{11+5{\cos }^2\theta }{\cos }^2\theta +16-8\cos \theta =11+5{\cos }^2\theta 4{\cos }^2\theta +8\cos \theta -5=0(2\cos \theta -1)(2\cos \theta +5)=0\cos \theta =\frac{1}{2}\Rightarrow \theta =\pm \frac{\pi }{3}$

Hence, the points $P=(2,\pm 8\sqrt{\frac{3}{11}})$