MathematicsMedium74×since 2004Q3756Let x2a2+y2b2=1{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1a2x2+b2y2=1 (a > b) be a given ellipse, length of whose latus rectum is 10. If its eccentricity is the maximum value of the function, ϕ(t)=512+t−t2\phi \left( t \right) = {5 \over {12}} + t - {t^2}ϕ(t)=125+t−t2, then a² + b² is equal to :A145B126C135D116Check answerSkip