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

We are asked to determine $$a^2 + b^2$$ for an ellipse whose latus rectum has length 30 and whose eccentricity equals the maximum value of the function $$f(t) = -\f\frac{3}{4} + 2t - t^2\,. $$

Completing the square gives $$f(t) = -\bigl(t^2 - 2t + 1\bigr) + 1 - \f\frac{3}{4} = -(t - 1)^2 + \f\frac{1}{4},$$ so the maximum value of $$f(t)$$ is $$\tf\frac{1}{4}$$, attained at $$t = 1$$. Equivalently, one may note $$f'(t) = 2 - 2t = 0 \implies t = 1,\qquad f(1) = -\f\frac{3}{4} + 2 - 1 = \f\frac{1}{4},$$ and conclude that the eccentricity of the ellipse is $$e = \tf\frac{1}{4}\,. $$

The length of the latus rectum of an ellipse is given by $$\f\frac{2b^2}{a}\,. $$ Since this equals 30, we have $$\f\frac{2b^2}{a} = 30 \implies b^2 = 15a\,.$$

On the other hand, the relation between the semi-axes and the eccentricity is $$b^2 = a^2\bigl(1 - e^2\bigr) = a^2\Bigl(1 - \f\frac{1}{16}\Bigr) = \f\frac{15a^2}{16}\,. $$ Equating this to $$15a$$ yields $$\f\frac{15a^2}{16} = 15a \implies a = 16\,, $$ and hence $$b^2 = 15 \t\times 16 = 240\,.$$

Finally, $$a^2 + b^2 = 256 + 240 = 496\,, $$ so the required value is Option 3: 496.