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

$\text{ Equation of tangent of }\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\text{ with slope }m\text{ is }$

$\begin{array}{r}y=mx\pm \sqrt{a^2{m}^2+b^2}\\ \Rightarrow y=2x\pm \sqrt{4a^2+b^2}\end{array}$

$\text{ It is normal to the circle }{x}^2+y^2+4x+1=0\text{ hence, this tangent }$

$\begin{array}{l}\text{ passes through centre }(-2,0)\\ \Rightarrow 0=-4\pm \sqrt{4a^2+b^2}\\ \Rightarrow 4a^2+b^2=16\end{array}$

$\text{ Using Arithmetic mean (A.M.) }\ge \text{ Geometric mean (G.M.) }$

$\begin{array}{l}\text{ we get }\frac{4a^2+b^2}{2}\ge \sqrt{4a^2{b}^2}\\ \Rightarrow ab\le 4\end{array}$

$$\begin{gathered} \text{ Minimum value of ab is 4 } \\ \text{Therefore, the correct answer is 0004.00} \end{gathered}$$