Let x2a2+y2b2=1 ,(a>b) be given ellipse, length of whose latus rectum – InterviewSolution

1.	Let x2a2+y2b2=1 ,(a>b) be given ellipse, length of whose latus rectum is 10. If its eccentricity is the maximum value of the function, ϕ(t)=512+t−t2, then a2+b2 is equal to
Answer» Let $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, (a > b)$ be given ellipse, length of whose latus rectum is $10.$ If its eccentricity is the maximum value of the function, $ϕ (t) = \frac{5}{12} + t - t^{2}, then a^{2} + b^{2}$ is equal to

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