Question

If the tangent at the point $(4cos\varphi ,\frac{16}{\sqrt{11}}sin\varphi )$to the ellipse $16x^2+11y^2=256$ is also a tangent to the circle $(x-1{)}^2+y^2=4^2$ then the value of $\varphi$ is

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

Answer 1

The correct option is C $\pm \frac{\pi }{3}$

Equation of tangent at $(4cos\varphi ,\frac{16}{\sqrt{11}}sin\varphi )$ is $4xcos\varphi +y\sqrt{11}sin\varphi =16$
Distance from this tangent to center should be radius of circle.
$(x-1{)}^2+y^2=4^2,cp=r\Rightarrow \left|\frac{4cos\varphi -16}{\sqrt{16co{s}^2\varphi +11si{n}^2\varphi }}\right|=4$
$4co{s}^2\varphi +8cos\varphi -5=0\Rightarrow (2cos\varphi -1)(2cos\varphi +5)=0$
$\therefore cos\varphi =\frac{1}{2},\varphi =\pm \frac{\pi }{3}$