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 of point of contact is

• A. ${45}^{\circ }$
• B. ${90}^{\circ }$
• C. ${30}^{\circ }$
• D. $0^{\circ }$

Answer 1

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

Let $\theta$ be the eccentric angle of the point of contact, then tangent at this point is $\frac{x\cos \theta }{a}+\frac{y\sin \theta }{b}=1$
Comparing it with given equation of line $\frac{x}{a\sqrt{2}}+\frac{y}{b\sqrt{2}}=1,$ we get
$\Rightarrow \frac{\cos \theta }{\frac{1}{\sqrt{2}}}=\frac{\sin \theta }{\frac{1}{\sqrt{2}}}=1$
$\Rightarrow \cos \theta =\frac{1}{\sqrt{2}}$ and $\sin \theta =\frac{1}{\sqrt{2}}$
$\Rightarrow \theta ={45}^{\circ }$