Question

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

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

Answer 1

Answer: (A), (B)

The correct options are
A the maximum value of $ab$ is $\frac{2}{3}$
B $a\in (\sqrt{\frac{2}{5}},2)$

$y=\frac{1}{3}x\pm \sqrt{\frac{a^2}{9}+b^2},$ it passes
through $(-1,-1)$
$\Rightarrow -3=-1\pm \sqrt{a^2+9b^2}$
$\Rightarrow a^2+9b^2=4$

A.M. $\ge$ G.M.
$\Rightarrow \frac{4}{2}\ge a\cdot 3b$
$\Rightarrow ab\le \frac{2}{3}$

$a^2+9a^2(1-e^2)=4$
$\Rightarrow e^2=\frac{10a^2-4}{9a^2}$
$0<e^2<1\Rightarrow a\in (\sqrt{\frac{2}{5}},2)$