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 $f\left(t\right)=−\frac{3}{4}+2t−{t}^{2}$, then $\left(a^2+b^2\right)$ is equal to

Question

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 $f\left(t\right)=−\frac{3}{4}+2t−{t}^{2}$, then $\left(a^2+b^2\right)$ is equal to

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Let the length of the latus rectum of an ellipse \(\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1\), \((a > b)\), …

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Let the length of the latus rectum of an ellipse \(\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1\), \((a &gt; b)\), …

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Let the length of the latus rectum of an ellipse \(\frac{{x}^{2}}{{a}^{2}}+\frac{{y}^{2}}{{b}^{2}}=1\), \((a > b)\), …