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 $f(t) = t^2 + t + \frac{11}{12}$, $t \in \mathbb{R}$, then $a^2 + b^2$ is equal to :

<p>Given that the length of the latus rectum of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is 10, we have:</p> <p>$ \frac{2b^2}{a} = 10 \quad \Rightarrow \quad 5a = b^2 \tag{1} $</p> <p>Next, consider the function $ f(t) = t^2 + t + \frac{11}{12} $. To find its minimum value, we calculate the derivative:</p> <p>$ \frac{df(t)}{dt} = 2t + 1 = 0 \quad \Rightarrow \quad t = \frac{-1}{2} $</p> <p>Plugging $t = -\frac{1}{2}$ into $f(t)$ gives the minimum value:</p> <p>$ f\left(-\frac{1}{2}\right) = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) + \frac{11}{12} = \frac{1}{4} - \frac{1}{2} + \frac{11}{12} = \frac{3 - 6 + 11}{12} = \frac{8}{12} = \frac{2}{3} $</p> <p>Thus, the eccentricity $e$ of the ellipse is $\frac{2}{3}$, so $e^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9}$.</p> <p>Using the eccentricity formula for an ellipse:</p> <p>$ e^2 = \frac{1 - b^2}{a^2} \quad \Rightarrow \quad \frac{4}{9} = \frac{1 - b^2}{a^2} $</p> <p>Rearranging gives:</p> <p>$ b^2 = a^2 \left(1 - \frac{4}{9}\right) = a^2 \cdot \frac{5}{9} $</p> <p>From equation $(1)$, $b^2 = 5a$. Substituting, we have:</p> <p>$ 5a = a^2 \cdot \frac{5}{9} \quad \Rightarrow \quad a^2 = 9a $</p> <p>Solving for $a$,</p> <p>$ a = 9, \quad \text{and therefore } b^2 = 5a = 45 \quad \Rightarrow \quad b = \sqrt{45} = 3\sqrt{5} $</p> <p>Finally, calculate $a^2 + b^2$:</p> <p>$ a^2 + b^2 = 81 + 45 = 126 $</p>