If a tangent of slope 2 to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ touches the circle $x^2+y^2=4$,…
If a tangent of slope 2 to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ touches the circle $x^2+y^2=4$, then maximum value of $a b$ is
4
12
5
7
Solution
Given the ellipse, $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Since, tangent to the given ellipse with slope $m=2$ is given as $y=m x \pm \sqrt{a^2 m^2+b^2} \Rightarrow y=2 x \pm \sqrt{4 a^2+b^2}$
$\because$ it touches the circle $x^2+y^2=4$
So, $2=\frac{\sqrt{4 a^2+b^2}}{\sqrt{1+4}}$ [distance from origin to the tangent]
$\begin{aligned}
& \Rightarrow \sqrt{4 a^2+b^2}=2 \sqrt{5} \Rightarrow 4 a^2+b^2=20 \\
& \because A \cdot M . \geq G \cdot M . \Rightarrow \frac{4 a^2+b^2}{2} \geq \sqrt{4 a^2 b^2} \\
& \Rightarrow 10 \geq 2 a b \Rightarrow a b \leq 5
\end{aligned}$
So, maximum value of $a b=5$.