Let ${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$ (a > b) be a given ellipse, length of whose latus rectum is 10. If its eccentricity is the maximum value of the function,
$\phi \left( t \right) = {5 \over {12}} + t - {t^2}$, then a² + b² is equal to :

Given ellipse ${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$ (a &gt; b)<br><br>Length of latus rectum $= {{2{b^2}} \over a} = 10$<br><br>$\phi (t) = {5 \over {12}} + t - {t^2}$<br><br>$= {8 \over {12}} - {\left( {t - {1 \over 2}} \right)^2}$<br><br>$\therefore$ $\phi {(t)_{\max }} = {8 \over {12}} = {2 \over 3}$<br><br>$\therefore$ eccentricity (e) = ${2 \over 3}$<br><br>Also, ${e^2} = 1 - {{{b^2}} \over {{a^2}}}$<br><br>$\Rightarrow {4 \over 9} = 1 - {{{b^2}} \over {{a^2}}}$<br><br>$\Rightarrow {{{b^2}} \over {{a^2}}} = {5 \over 9}$<br><br>$\Rightarrow {{{b^2}} \over a} \times {1 \over a} = {5 \over 9}$<br><br>$\Rightarrow {5 \over a} = {5 \over 9}$<br><br>$\Rightarrow a = 9$<br><br>$\therefore$ ${b^2} = 5 \times 9 = 45$<br><br>$\therefore$ ${a^2} + {b^2} = 81 + 45 = 126$