Question

A normal is drawn to the ellipse $\frac{x^2}{(a^2+4a+5{)}^2}+\frac{y^2}{(a^2+4{)}^2}=1$ whose center is at origin $O$. If the maximum radius of the circle ,centered at the origin and touching the normal is $25$ then the positive value of $a$ is

Answer 1

The equation of the normal at $(\alpha \cos \theta ,\beta \sin \theta )$ to the ellipse $\frac{x^2}{{\alpha }^2}+\frac{y^2}{{\beta }^2}=1$
(here,$\alpha =(a^2+4a+5),\beta =(a^2+4)$) is $\alpha x\sec \theta -\beta y\text{cosec}\theta ={\alpha }^2-{\beta }^{2}r=\frac{|{\alpha }^2-{\beta }^2|}{\sqrt{{\alpha }^2{\sec }^2\theta +{\beta }^2{\text{cosec}}^2\theta }}=\frac{|{\alpha }^2-{\beta }^2|}{\sqrt{{\alpha }^2+{\beta }^2+{\alpha }^2{\tan }^2\theta +{\beta }^2{\cot }^2\theta }}{r}_{max}=\frac{|{\alpha }^2-{\beta }^2|}{\sqrt{{\alpha }^2+{\beta }^2+2\alpha \beta }}\left\{\text{Using AM > GM}\right\}=|\alpha -\beta |=|4a+1|r_{max}=|4a+1|=25a=6$