Question

If a tangent of slope $2$ of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is normal to the circle $x^2+y^2+4x+1=0$, then the maximum value of $ab$ is

• A. $4$
• B. $2$
• C. $1$
• D. $Noneofthese$

Answer 1

The correct option is A $4$

A tangent of slope $2$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is

$y=2x\pm \sqrt{4a^2+b^2}$ (i)

This is normal to the circle $x^2+y^2+4x+1=0$

$\Rightarrow$ Equation (i), passes through center of the circle$(-2,0)$

$\Rightarrow 0=-4\pm \sqrt{4a^2+b^2}$

$\Rightarrow 4a^2+b^2=16$

Usin A.M $\ge$ G.M., we get

$\frac{4a^2+b^2}{2}\ge \sqrt{4a^2{b}^2}$

$\Rightarrow ab\le 4$

$\therefore$ the minimum value of $ab$ is $4$