Let the length of a latus rectum of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ be 10 . If its eccentricity is the minimum value of the function $f(\mathrm{t})=\mathrm{t}^2+\mathrm{t}+\frac{11}{12}$, $\mathrm{t} \in \mathbf{R}$, then $\mathrm{a}^2+\mathrm{b}^2$ is equal to :

Question

Aicharya · Accepted Answer

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

Let the length of a latus rectum of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ be 10 . If its…

Solution