Question

If $(\sqrt{3})bx+ay-2ab$ 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. $\pi /6$
• B. $\pi /4$
• C. $\pi /3$
• D. $\pi /2$

Answer 1

Answer: (A)

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