Question

If $\left(\sqrt{3}\right)bx+ay=2ab$ touches the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, then the eccentric angle of the point of contact is

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

Answer 1

Answer: (A)

$\frac{\pi }{6}$

Equation of tangent $\frac{x}{a}\frac{\sqrt{3}}{2}+\frac{y}{b}\frac{1}{2}=1$ (i)
and equation of tangent at the point $(a\cos \varphi ,b\sin \varphi )$ is
$\frac{x}{a}\cos \varphi +\frac{y}{b}\sin \varphi =1$ (ii)
Comparing (i) and (ii), we have $\cos \varphi =\frac{\sqrt{3}}{2}$ and $\sin \varphi =\frac{1}{2}$
Hence, $\varphi =\frac{\pi }{6}$