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

Answer 1

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

$\frac{x}{a}+\frac{y}{b}=\sqrt{2}$ is tangent of ellipse

Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$ & $(x_1,y_1)$ is at ellipse

tangent of an ellipse is -

$\frac{x{x}_1}{a^2}+\frac{y{y}_1^2}{b^2}=1$

Let $(acos\theta ,bsin\theta )$ is at ellipse, then

tangent $\frac{xcos\theta }{a}+\frac{ysin\theta }{b}=1$ _____ (1)

$\frac{x}{a\sqrt{2}}+\frac{y}{b\sqrt{2}}=1$ ____ (2) given in question

compare (1) & (2)

$cos\theta =\frac{1}{\sqrt{2}}$ co-efficient of x.

$cos\theta =cos{45}^{\circ }$

$\theta ={45}^{\circ }$