Question

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

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

Answer 1

Answer: (C)

${45}^0$

Equation of any tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is

$\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1$ ...(1)

Also, $\frac{x}{a}+\frac{y}{b}=\sqrt{2}$ ....(2)

touches the given ellipse.

Comparing coefficients in (1) and (2), we get

$\frac{\left(\cos \theta \right)/a}{1/a}=\frac{\left(\sin \theta \right)/b}{1/b}=\frac{1}{\sqrt{2}}$

$\therefore \cos \theta =\frac{1}{\sqrt{2}}=\sin \theta \Rightarrow \theta ={45}^0$

Hence, option C.