Mathematics - Ellipse Question with Solution | TestHub

Length of LR $=\frac{2b^2}{a}=10\Rightarrow 5\mathrm{a}={\mathrm{b}}^2$...(1)
$\begin{array}{rl}& \mathrm{f}(\mathrm{t})={\mathrm{t}}^2+\mathrm{t}+\frac{11}{12}\\ & \frac{\mathrm{df}(\mathrm{t})}{\mathrm{dt}}=2\mathrm{t}+1=0\Rightarrow \mathrm{t}=\frac{-1}{2}\\ & \text{ Min value of }\mathrm{f}(\mathrm{t})={\left(\frac{-1}{2}\right)}^2+\left(\frac{-1}{2}\right)+\frac{11}{12}\\ & =\frac{1}{4}\frac{-1}{2}+\frac{11}{12}=\frac{3-6+11}{12}=\frac{8}{12}=\frac{2}{3}=\mathrm{e}\\ & {\mathrm{e}}^2=\frac{1-{\mathrm{b}}^2}{{\mathrm{a}}^2}\Rightarrow \frac{4}{9}=\frac{1-{\mathrm{b}}^2}{{\mathrm{a}}^2}\end{array}$
$\Rightarrow \frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}=\frac{1-4}{\mathrm{a}}=\frac{5}{\mathrm{a}}\Rightarrow {\mathrm{b}}^2=\frac{5{\mathrm{a}}^2}{\mathrm{a}}$...(2)
$\begin{array}{rl}& \text{ From (1) \& (2) }\\ & 5\mathrm{a}=\frac{5{\mathrm{a}}^2}{\mathrm{a}}\Rightarrow \mathrm{a}=9,\mathrm{b}=\sqrt{45}=3\sqrt{5}\\ & \therefore \mathrm{a}2+\mathrm{b}2=81+45=126\end{array}$