Question

If a tangent of slope $\frac{1}{3}$ of the ellipse $\frac{x^3}{a^2}+\frac{y^2}{b^2}=1(a>b)$ is normal to the circle $x^2+y^2+2x+2y+1=0$ then

• A. maximum value of $ab$ is $\frac{2}{3}$
• B. $a\in \left(\sqrt{\frac{2}{5}}2\right)$
• C. $a\in (\frac{2}{5},2)$
• D. maximum value of $ab$ is $1$

Answer 1

The correct option is A maximum value of $ab$ is $\frac{2}{3}$

Given that,

$\begin{array}{c}\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\\ y=mx-\sqrt{a^2{m}^2+b^2}\end{array}$

Equation of tangent.

As tangent is normal to circle $x^2+y^2+2x-y+1=0$

$\therefore$ it passes three center of circle.

i.e. $(-1,-1)$

$\begin{array}{c}-1=-m-\sqrt{a^2{m}^2+b^2}m=\frac{1}{3}\\ -1=-\frac{1}{3}-\sqrt{\frac{a^2}{9}+b^2}\\ \sqrt{\frac{a^2}{9}+b^2}=\frac{2}{3}\\ \Rightarrow \frac{a^2}{9}+b^2=\frac{4}{9}\\ \Rightarrow a^2+a{b}^2=4\\ A.M\ge G.M\\ \Rightarrow \frac{a^2+a{b}^2}{2}\ge \sqrt{a^2\times a{b}^2}\\ \Rightarrow \frac{2}{3}\ge ab\\ ab\le \frac{2}{3}\end{array}$

$\therefore$ max. value of $ab=\frac{2}{3}$

Hence,

Option $A$ is correct answer.