Question

lf $\left(\sqrt{3}\right)bx+ay=2ab$ is tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, then eccentric angle $\theta$ is

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

Answer 1

The correct option is B $\frac{\pi }{6}$

Let eccentric angle be $\theta$ then the point is $(acos\theta ,bsin\theta )$

Slope of a tangent at a point $\theta$ is given by $-\frac{b}{a}\cot \theta$

Slope of the line given in question is $-\frac{b}{a}\sqrt{3}$

Equating both we get $\theta$ = $\frac{\pi }{6}$