Question

If the tangent to the ellipse $x^2+4y^2=16$ at the point ${}^{\prime }{\varphi }^{\prime }$ is a normal to the circle $x^2+y^2-8x-4y=0$, then $\varphi$ is equal to

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

Answer 1

Answer: (A)

$\frac{\pi }{2}$

Given ellipse may be written as, $\frac{x^2}{16}+\frac{y^2}{4}=1$
Now tangent on this ellipse at any point ${}^{\prime }{\varphi }^{\prime }$ is given by,
$\frac{x\sin \varphi }{4}+\frac{y\cos \varphi }{2}=1$
Also this line is normal to the given circle, so it will pass through centre of the circle which is $(4,2)$
$\Rightarrow \cos \varphi +\sin \varphi =1$
Squaring we get, $\sin 2\varphi =0\Rightarrow \varphi =\frac{\pi }{2}$