Question

If $x+y\sqrt{2}=2\sqrt{2}$ is a tangent to the ellipse $x^2+2y^2=4,$ then the eccentric angle of the point of contact is

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is B $\frac{\pi }{4}$

Given, equation of ellipse is
$\frac{x^2}{4}+\frac{y^2}{2}=1$
$\Rightarrow a=2,b=\sqrt{2}$
Equation of tangent is
$x+y\sqrt{2}=2\sqrt{2}$
$\Rightarrow \frac{x}{2}\cdot \frac{1}{\sqrt{2}}+\frac{y}{\sqrt{2}}\cdot \frac{1}{\sqrt{2}}=1$

Equation of tangent to the ellipse with eccentric angle $\theta$ is given by,
$\frac{x\cos \theta }{a}+\frac{y\sin \theta }{b}=1$
$\Rightarrow \frac{x\cos \theta }{2}+\frac{y\sin \theta }{\sqrt{2}}=1$
Now, comparing this with given equation we get,
$\cos \theta =\sin \theta =\frac{1}{\sqrt{2}}$ ​
Hence, required eccentric angle is, $\theta =\frac{\pi }{4}$