Question

If $\frac{x}{a}+\frac{y}{b}=\sqrt{2}$ touches the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$ then its eccentric angle $\theta$ is equal to

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

Answer 1

The correct option is C ${45}^{\circ }$

Equation of the tangent at point $\theta$ is $\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1....\left(1\right)$
Given line is $\frac{x}{a}.\frac{1}{\sqrt{2}}+\frac{y}{b}.\frac{1}{\sqrt{2}}=1....\left(2\right)$
Since (2) touches the ellipse, (1), (2) represent the same line.
$\because \cos \theta =\frac{1}{\sqrt{2}},\sin \theta =\frac{1}{\sqrt{2}}$
$\therefore \theta ={45}^{\circ }$