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 $\theta$ 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)

Let point of contact be $(a\cos \theta ,b\sin \theta )$
$\therefore$ equation of tangent be $\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1$
equation of given line is $\frac{x}{a}\cdot \frac{\sqrt{3}}{2}+\frac{y}{b}\cdot \frac{1}{2}=1$
$\therefore \cos \theta =\frac{\sqrt{3}}{2}$ and $\sin \theta =\frac{1}{2}$
$\Rightarrow \theta =\frac{\pi }{6}$