Question

Angle between the tangents drawn from point $(4,5)$ to the ellipse $\frac{x^2}{16}+\frac{y^2}{25}=1$ is

• A. $\frac{\pi }{3}$
• B. $\frac{5\pi }{6}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is D $\frac{\pi }{2}$

The angle between two tangents of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ drawn from the point $(x_0,y_0)$ is given by

$\tan \theta =\frac{2\sqrt{a^2{y}_0^2+b^2{x}_0^2-a^2{b}^2}}{x_0^2+y_0^2-a^2-b^2}$

Here, $a^2=16,b^2=25,x_0=4$ and $y_0=5$.

$\tan \theta =\frac{2\sqrt{\left(16\right)\left(25\right)+\left(25\right)\left(16\right)-\left(16\right)\left(25\right)}}{16+25-16-25}$

$\tan \theta =\frac{2\sqrt{400}}{0}$

$\tan \theta =\infty$

$\theta ={\tan }^{-1}\left(\infty \right)$

$\therefore$ $\theta =\frac{\pi }{2}$