Question

If a tangent of slope $3$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is normal to the circle $2x^2+2y^2+8x+1=0,$ then the maximum value of $ab$ is

Answer 1

Since, the line having slope $3$ is normal to the circle $2x^2+2y^2+8x+1=0,$ then it must pass through the center of the circle i.e., $(-2,0)$

So, the equation of the tangent is,
$y-0=3(x+2)$
$\Rightarrow y=3x+6$
Comparing with the general equation of the tangent $y=mx\pm \sqrt{a^2{m}^2+b^2}$, we have
$m=3$ and $a^2{m}^2+b^2=36$
$\Rightarrow 9a^2+b^2=36...\left(1\right)$

Using A.M.-G.M. inequality
$\frac{9a^2+b^2}{2}\ge \sqrt{9a^2\times b^2}$
$\Rightarrow \frac{36}{2}\ge 3|ab|$ [From eqn$\left(1\right)$]
$\Rightarrow |ab|\le 6$
$\Rightarrow -6\le ab\le 6$

$\therefore$ Maximum value of $ab=6$