Thank you for reporting, we will resolve it shortly
Q.
If $\sqrt{3} b x+a y=2 a b$ 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
Conic Sections
Solution:
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}$