A normal is drawn to the ellipse $\frac{{\mathrm{x}}^2}{{\left({\mathrm{a}}^2+2\mathrm{a}+2\right)}^2}+\frac{{\mathrm{y}}^2}{{\left({\mathrm{a}}^2+1\right)}^2}=1$whose centre is at O. If maximum radius of the circle, centered at the origin and touching the normal, is  5 then the positive value of 'a' is.......

Equation of the ellipse $\frac{{\mathrm{x}}^2}{{\left({\mathrm{a}}^2+2\mathrm{a}+2\right)}^2}+\frac{{\mathrm{y}}^2}{{\left({\mathrm{a}}^2+1\right)}^2}=1$

put $\alpha =a^2+2a+2,\beta =a^2+1$

$\Rightarrow$ It is in the form of $\frac{{\mathrm{x}}^2}{{\alpha }^2}+\frac{{\mathrm{y}}^2}{{\beta }^2}=1$

$$\begin{gathered} equation of normal at (\alpha \cos \theta ,\beta \sin \theta ) is \alpha xsec\theta -\beta ycosec\theta ={\alpha }^2-{\beta }^2 \\ \sin ce it touches the circle then r=d \\ r=\frac{|{\alpha }^2-{\beta }^2|}{\sqrt{{\alpha }^{2}se{c}^2\theta +{\beta }^2\cos e{c}^2\theta }} \\ =\frac{|{\alpha }^2-{\beta }^2|}{\sqrt{{\alpha }^2+{\beta }^2+{\alpha }^2{\tan }^2\theta +{\beta }^{2}co{t}^2\theta }} \\ {r}_{max}=\frac{|{\alpha }^2-{\beta }^2|}{\sqrt{{\alpha }^2+{\beta }^2+2\alpha \beta }}(usingAM>GM) \\ =|\alpha -\beta | \\ =|2a+1| \\ {r}_{max}=|2a+1|=5 \\ \Rightarrow a=2 \end{gathered}$$