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

Answer: (C)

Equation of any tangent to the ellipse
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is
$\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =\mathrm{1...}$(1)
Also, $\frac{x}{a}+\frac{y}{b}=\sqrt{2}$ touches the given ellipse.
Comparing coefficients in (1) and (2), we get
$\frac{\cos \theta /a}{1/a}=\frac{\sin \theta /b}{1/b}=\frac{1}{\sqrt{2}}$
$\Rightarrow \cos \theta =\frac{1}{\sqrt{2}}=\sin \theta$
$\therefore \theta =45^{\circ }$