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(B) The length of the latus rectum of the ellipse is given by $\frac{2b^2}{a} = 10$,which implies $b^2 = 5a$. To find the eccentricity $e$,we minimize

Vedclass · Accepted Answer

$126$

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(B) The length of the latus rectum of the ellipse is given by $\frac{2b^2}{a} = 10$,which implies $b^2 = 5a$. To find the eccentricity $e$,we minimize $f(t) = t^2 + t + \frac{11}{12}$. Taking the derivative,$f'(t) = 2t + 1 = 0$,so $t = -\frac{1}{2}$. The minimum value is $f(-\frac{1}{2}) = (-\frac{1}{2})^2 - \frac{1}{2} + \frac{11}{12} = \frac{1}{4} - \frac{1}{2} + \frac{11}{12} = \frac{3 - 6 + 11}{12} = \frac{8}{12} = \frac{2}{3}$. Thus,$e = \frac{2}{3}$,so $e^2 = \frac{4}{9}$. For an ellipse,$e^2 = 1 - \frac{b^2}{a^2}$,so $\frac{4}{9} = 1 - \frac{b^2}{a^2}$,which gives $\frac{b^2}{a^2} = 1 - \frac{4}{9} = \frac{5}{9}$. Substituting $b^2 = 5a$,we get $\frac{5a}{a^2} = \frac{5}{9}$,so $\frac{5}{a} = \frac{5}{9}$,which implies $a = 9$. Then $b^2 = 5(9) = 45$. Therefore,$a^2 + b^2 = 9^2 + 45 = 81 + 45 = 126$.

Let the length of a latus rectum of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ be $10$. If its eccentricity is the minimum value of the function $f(t) = t^2 + t + \frac{11}{12}$,$t \in R$,then $a^2 + b^2$ is equal to:

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