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

Question

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 \(f(\mathrm{t})=\mathrm{t}^{2}+\mathrm{t}+\frac{11}{12}, t \in \mathbf{R},\) then \(\mathrm{a}^{2}+\mathrm{b}^{2}\) is equal to :

(1) 125

(2) 126

(3) 120

(4) 115 

Answer 1

Correct option is: (2) 126 

Length of \(\mathrm{LR}=\frac{2 \mathrm{~b}^{2}}{\mathrm{a}}=10 \Rightarrow 5 \mathrm{a}=\mathrm{b}^{2} \quad\cdots(1)\)

\(\mathrm{f}(\mathrm{t})=\mathrm{t}^{2}+\mathrm{t}+\frac{11}{12} \)

\(\frac{\mathrm{df}(\mathrm{t})}{\mathrm{dt}}=2 \mathrm{t}+1=0 \Rightarrow \mathrm{t}=\frac{-1}{2}\)

Min value of \(f(t)=\left(\frac{-1}{2}\right)^{2}+\left(\frac{-1}{2}\right)+\frac{11}{12}\)

\(=\frac{1}{4} \frac{-1}{2}+\frac{11}{12}=\frac{3-6+11}{12}=\frac{8}{12}=\frac{2}{3}=\mathrm{e}\)

\(\mathrm{e}^{2}=\frac{1-\mathrm{b}^{2}}{\mathrm{a}^{2}} \Rightarrow \frac{4}{9}=\frac{1-\mathrm{b}^{2}}{\mathrm{a}^{2}}\)

\(\begin{equation*} \Rightarrow \frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}=\frac{1-4}{\mathrm{a}}=\frac{5}{\mathrm{a}} \Rightarrow \mathrm{b}^{2}=\frac{5 \mathrm{a}^{2}}{\mathrm{a}}\quad \cdots(2) \end{equation*}\)

From (1) \(\&\) (2)

\(5 \mathrm{a}=\frac{5 \mathrm{a}^{2}}{\mathrm{a}} \Rightarrow \mathrm{a}=9, \quad \mathrm{~b}=\sqrt{45}=3 \sqrt{5}\)

\(\therefore\ \mathrm{a} 2+\mathrm{b} 2=81+45=126\)