Let the length of a latus rectum of an ellipse x^2 a^2+ y^2 b^2=1 be 10 . If its eccentricity is the minimum value of the function f( t)= t^2+ t+ 11…

Length of LR $=\frac{2 b^2}{a}=10 \Rightarrow 5 \mathrm{a}=\mathrm{b}^2$=a2b2​=10⇒5a=b2...(1)

$$\begin{aligned} & \mathrm{f}(\mathrm{t})=\mathrm{t}^2+\mathrm{t}+\frac{11}{12} \\ & \frac{\mathrm{df}(\mathrm{t})}{\mathrm{dt}}=2 \mathrm{t}+1=0 \Rightarrow \mathrm{t}=\frac{-1}{2} \\ & \text { Min value of } \mathrm{f}(\mathrm{t})=\left(\frac{-1}{2}\right)^2+\left(\frac{-1}{2}\right)+\frac{11}{12} \\ & =\frac{1}{4} \frac{-1}{2}+\frac{11}{12}=\frac{3-6+11}{12}=\frac{8}{12}=\frac{2}{3}=\mathrm{e} \\ & \mathrm{e}^2=\frac{1-\mathrm{b}^2}{\mathrm{a}^2} \Rightarrow \frac{4}{9}=\frac{1-\mathrm{b}^2}{\mathrm{a}^2}\end{aligned}$$​f(t)=t2+t+1211​dtdf(t)​=2t+1=0⇒t=2−1​ Min value of f(t)=(2−1​)2+(2−1​)+1211​=41​2−1​+1211​=123−6+11​=128​=32​=ee2=a21−b2​⇒94​=a21−b2​​

$\Rightarrow \frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{1-4}{\mathrm{a}}=\frac{5}{\mathrm{a}} \Rightarrow \mathrm{b}^2=\frac{5 \mathrm{a}^2}{\mathrm{a}}$⇒a2b2​=a1−4​=a5​⇒b2=a5a2​...(2)

$$\begin{aligned} & \text { From (1) \& (2) } \\ & 5 \mathrm{a}=\frac{5 \mathrm{a}^2}{\mathrm{a}} \Rightarrow \mathrm{a}=9, \quad \mathrm{~b}=\sqrt{45}=3 \sqrt{5} \\ & \therefore \mathrm{a} 2+\mathrm{b} 2=81+45=126\end{aligned}$$​ From (1) & (2) 5a=a5a2​⇒a=9, b=45​=35​∴a2+b2=81+45=126​