Question

lf $x+y\sqrt{2}=2\sqrt{2}$ 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

Answer: (B)

$\frac{\pi }{4}$

Equation of tangent to the ellipse with eccentric angle $\theta$ is given by,
$\frac{x\cos \theta }{a}+\frac{y\sin \theta }{b}=1$
Now comparing this with given equation we get, $\cos \theta =\sin \theta =\frac{1}{\sqrt{2}}$
since $a=2,b=\sqrt{2}$
Hence required eccentric angle is, $\theta ={45}^o$