Question

Shortest distance of curve $2x^2+5x{y}^{}+2y^2=1$ from origin is

• A. $2$
• B. $5$
• C. $3$
• D. None of these

Answer 1

The correct option is A $2$

${2x}^2+5xy+2y^2=1$

Let $x=r\cos \theta$, $y=r\sin \theta$ be any point on curve.

$\Rightarrow {2r}^2{\cos }^2\theta +5.r\cos \theta .r\sin \theta +2r^2{\sin }^2\theta =1$

$\Rightarrow {2r}^2({\cos }^2\theta +{\sin }^2\theta )+5r^2\left(\sin \theta \cos \theta \right)=1$

$\Rightarrow {2r}^2+{5r}^2\sin \theta \cos \theta =1$

$\Rightarrow r^2[2+\frac{5}{2}\sin 2\theta ]=1$

$\Rightarrow r^2=\frac{1}{2+5/2\sin 2\theta }$

$r$ is minimum if $2+\frac{5}{2}\sin 2\theta$ is max $=2+\frac{5}{2}=\frac{9}{2}$

$\therefore$ $r_{min}^2=\frac{1}{9/2}\Rightarrow r_{min}=\sqrt{\frac{2}{9}}$ [D]