Let the length of the latus rectum of an elli | vedclass

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(C) The function is $f(t) = -t^{2} + 2t - \frac{3}{4}$.To find the maximum value, we complete the square: $f(t) = -(t^{2} - 2t + 1) + 1 - \frac{3}{4}

Vedclass · Accepted Answer

$496$

Vedclass · Answer

(C) The function is $f(t) = -t^{2} + 2t - \frac{3}{4}$.To find the maximum value, we complete the square: $f(t) = -(t^{2} - 2t + 1) + 1 - \frac{3}{4} = -(t-1)^{2} + \frac{1}{4}$.The maximum value is $e = \frac{1}{4}$, so $e^{2} = \frac{1}{16}$.For an ellipse, $e^{2} = 1 - \frac{b^{2}}{a^{2}}$, so $\frac{b^{2}}{a^{2}} = 1 - \frac{1}{16} = \frac{15}{16} \Rightarrow b^{2} = \frac{15}{16}a^{2}$.The length of the latus rectum is $\frac{2b^{2}}{a} = 30$, so $b^{2} = 15a$.Equating the expressions for $b^{2}$: $\frac{15}{16}a^{2} = 15a$.Since $a 
eq 0$, we have $\frac{a}{16} = 1 \Rightarrow a = 16$.Then $b^{2} = 15(16) = 240$.Thus, $a^{2} + b^{2} = 16^{2} + 240 = 256 + 240 = 496$.

Let the length of the latus rectum of an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ $(a>b)$ be $30$. If its eccentricity is the maximum value of the function $f(t)=-\frac{3}{4}+2t-t^{2}$, then $(a^{2}+b^{2})$ is equal to -

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