Question
Let the length of the latus rectum of an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{{\text{ }b}^{2}} = 1,(a > b)$, be 30 . If its eccentricity is the maximum value of the function $f(t) = - \frac{3}{4} + 2t - t^{2}$, then $\left( a^{2} + b^{2} \right)$ is equal to -
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Solution
$f(t) = \frac{- 3}{4} + 2t - t^{2}$
$\left. \ f(t) \right|_{\text{maximum~}} = \frac{1}{4} = e \Rightarrow e^{2} = \frac{1}{16} \Rightarrow \frac{a^{2} - b^{2}}{a^{2}} = \frac{1}{16}$.
$$\begin{array}{r} \because\frac{2{\text{ }b}^{2}}{a} = 30 \Rightarrow {\text{ }b}^{2} = 15a\#(2) \end{array}$$
By (1) & (2)
$${16\left( a^{2} - 15a \right) = a^{2} \Rightarrow 15a^{2} - 16 \times 15a = 0 }{a = 16 }{b^{2} = 240 }{a^{2} + b^{2} = 256 + 240 }{= 496}$$
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