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 $\frac{\pi }{4}$

Any point $P\left(\theta \right)$ on the ellipse $\frac{x^2}{4}+\frac{y^2}{2}=1$can be written as $(2\cos \theta ,\sqrt{2}\sin \theta ).$

Tangent at $P:$
$\frac{x\cos \theta }{2}+\frac{y\sin \theta }{\sqrt{2}}=1\Rightarrow x\left(\sqrt{2}\cos \theta \right)+y\left(2\sin \theta \right)=2\sqrt{2}\dots \left(ii\right)$

Also, tangent at $P$ is given as
$x+y\sqrt{2}=2\sqrt{2}\dots \left(ii\right)$

If line $\left(i\right)$ and $\left(ii\right)$ are identical, then
$\frac{\sqrt{2}\cos \theta }{1}=\frac{2\sin \theta }{\sqrt{2}}\Rightarrow \tan \theta =1$
$\Rightarrow \theta =\frac{\pi }{4}$ (as $P$ lies in first quadrant)

Alternate Solution:
Given line can be rewritten as
$\frac{\sqrt{2}x}{4}+\frac{y}{2}=1$
and ellipse is $\frac{x^2}{4}+\frac{y^2}{2}=1$
Now using the point form of tangent, the point of contact will be $(\sqrt{2},1)$
Now from the parametric coordinates
$a\cos \theta =\sqrt{2},b\sin \theta =1\Rightarrow \cos \theta =\frac{1}{\sqrt{2}},\sin \theta =\frac{1}{\sqrt{2}}\Rightarrow \theta =\frac{\pi }{4}$