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

 

Slope of Tangent=2

 equation  of Tangent is $\mathrm{y}=2\mathrm{x}\pm \sqrt{4{\mathrm{a}}^2+{\mathrm{b}}^2} — ①$ 

centre $=(-2, 0)$

eq $①$ is normal to the circle then 

$$\begin{gathered} 0=-4\pm \sqrt{4{\mathrm{a}}^2+{\mathrm{b}}^2} \\ 4{\mathrm{a}}^2+{\mathrm{b}}^2=16 \\ \mathrm{A}.\mathrm{M}\ge \mathrm{G}.\mathrm{M} \\ \frac{4{\mathrm{a}}^2+{\mathrm{b}}^2}{2}\ge \sqrt{4{\mathrm{a}}^2.{\mathrm{b}}^2} \\ \frac{4{\mathrm{a}}^2+{\mathrm{b}}^2}{2}\ge 2\mathrm{ab} \\ \mathrm{ab} \le \frac{16}{4} \\ \Rightarrow ab\le 4 \end{gathered}$$