Question

lf $\frac{x}{a}+\frac{y}{b}=\sqrt{2}$ 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 equal to

• A. $0^0$
• B. ${90}^0$
• C. ${45}^0$
• D. ${60}^0$

Answer 1

The correct option is C ${45}^0$

Let any point on the ellipse be $P(a\cos \theta ,b\sin \theta )$

Since, the point lie on both straight line and ellipse

Substitute P in $\frac{x}{a}+\frac{y}{b}=\sqrt{2}$

$\Rightarrow \cos \theta +\sin \theta =\sqrt{2}$

$\Rightarrow \sin (\theta +\frac{\pi }{4})=1$
$\Rightarrow \theta +\frac{\pi }{4}=\frac{\pi }{2}$

Therefore, $\theta =\frac{\pi }{4}$