Question

If maximum distance of any point on the ellipse $x^2+2y^2+2xy=1$ from its centre be $r$, then $r$ is equal to:

• A. $3+\sqrt{3}$
• B. $2+\sqrt{2}$
• C. $\frac{\sqrt{2}}{\sqrt{3-\sqrt{5}}}$
• D. $\sqrt{2-\sqrt{2}}$

Answer 1

The correct option is C $\frac{\sqrt{2}}{\sqrt{3-\sqrt{5}}}$

Here centre of the ellipse is $(0,0)$

Let $P(r\cos \theta ,r\sin \theta )$ be any point on the given ellipse then $r^2{\cos }^2\theta +2r^2{\sin }^2\theta +2r^2\sin \theta \cos \theta =1$.

$\Rightarrow r^2=\frac{1}{{\cos }^2\theta +2{\sin }^2\theta +\sin 2\theta }$

$=\frac{1}{{\sin }^2\theta +1+\sin 2\theta }$

$=\frac{2}{1-\cos 2\theta +2+2\sin 2\theta }$

$=\frac{2}{3-\cos 2\theta +2\sin 2\theta }$

$\Rightarrow r_{max}=\frac{\sqrt{2}}{\sqrt{3-\sqrt{5}}}$