# Let $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} =1(a>b)$ be a given ellipse length of whose latus rectum is 10. If its eccentricity is the maximum value of the function, $\phi (t)=\frac{5}{12}+t-t^{2}$, then $a^{2}+b^{2}$ is equal to: Option: 1 145 Option: 2 116 Option: 3 126 Option: 4 135

$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b) ; \frac{2 b^{2}}{a}=10 \Rightarrow b^{2}=5 a\;\;\;\;\;\;\;\;\;\;\;\;\;\ldots(1)$

$\\\text { Now, } \phi(\mathrm{t})=\frac{5}{12}+\mathrm{t}-\mathrm{t}^{2}=\frac{8}{12}-\left(\mathrm{t}-\frac{1}{2}\right)^{2} \\ \phi(\mathrm{t})_{\max }=\frac{8}{12}=\frac{2}{3}=\mathrm{e} \Rightarrow \mathrm{e}^{2}=1-\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}=\frac{4}{9} \;\;\;\;\;\;\;\;\;\;\;\ldots(2)$

from (1) and (2)

$\\\Rightarrow \mathrm{a}^{2}=81\\ \text { So }, \mathrm{a}^{2}+\mathrm{b}^{2}=81+45=126$

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