Question

The minimum distance between the circle $x^2+y^2=9$ and the curve $2x^2+10y^2+6xy=1$ is

• A. $2\sqrt{2}$
• B. $2$
• C. $3-\sqrt{2}$
• D. $3-\frac{1}{\sqrt{11}}$

Answer 1

The correct option is B $2$

$x^2+y^2=9$

$2x^2+10y^2+6xy=1$

$\tan 2\theta =\frac{2h}{a-b}=\frac{6}{-8}=\frac{-3}{4}$

$\therefore \tan 2\theta =\frac{-3}{4}$

$\therefore \frac{2\tan \theta }{1-{\tan }^2\theta }=\frac{-3}{4}$

$\therefore (3\tan \theta +1)(\tan \theta -3)=0$

$\therefore \tan \theta =\frac{-1}{3},\tan \theta =3$

Let any $p$ on $2x^2+10y^2+6xy-1=0$ is $(r\cos \theta ,r\sin \theta )$

$\therefore 2r^2{\cos }^2\theta +10r^2{\sin }^2\theta +\theta r^2\sin \theta \cos \theta -1=0$

$=2r^2+10r^2{\tan }^2\theta +6r^2\tan \theta -1-{\tan }^2\theta =0$

$\therefore r^2=\frac{1+{\tan }^2\theta }{2+10{\tan }^2\theta +6\tan \theta }$

If $\tan \theta =3$

$\therefore r_1^2=\frac{1}{11}$

$\therefore r_1=\frac{1}{\sqrt{11}}$

If $\tan \theta =\frac{-1}{3},r_2^2=1\therefore r_2=1$

Hence, minimum distance $=3-1=2$