Question

If the tangent to the ellipse $x^2+4y^2=16$ at the point $P\left(\varphi \right)$ is normal to the circle $x^2+y^2-8x-4y=0,$ then possible values(s) of $\varphi$ is/are

• A. $\frac{\pi }{4}$
• B. $0$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (B), (D)

$\frac{\pi }{2}$

Tangent to ellipse at $P\left(\varphi \right)$ is $\frac{x}{4}\cos \varphi +\frac{y}{2}\sin \varphi =1.$

If this tangent is a normal to the circle $x^2+y^2-8x-4y=0,$ then it will pass through the centre$(4,2)$ of the circle, thus
$\frac{4}{4}\cos \varphi +\frac{2}{2}\sin \varphi =1$
$\Rightarrow \cos \varphi +\sin \varphi =1$
Squaring both the sides, we get
$\Rightarrow 1+\sin 2\varphi =1$
$\Rightarrow \sin 2\varphi =0$
$\Rightarrow 2\varphi =n\pi$
$\Rightarrow \varphi =\frac{n\pi }{2},n\in Z$

when $n=0\Rightarrow \varphi =0$
when $n=1\Rightarrow \varphi =\frac{\pi }{2}$