Let the length of a latus rectum of an ellipse (x^2)/(a^2) + (y^2)/(b^2) = 1 be 10.

Question

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 :

Accepted Answer

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:

$ \frac{2b^2}{a} = 10 \quad \Rightarrow \quad 5a = b^2 	ag{1} $

Next, consider the function $ f(t) = t^2 + t + \frac{11}{12} $. To find its minimum value, we calculate the derivative:

$ \frac{df(t)}{dt} = 2t + 1 = 0 \quad \Rightarrow \quad t = \frac{-1}{2} $

Plugging $t = -\frac{1}{2}$ into $f(t)$ gives the minimum value:

$ f\left(-\frac{1}{2}ight) = \left(-\frac{1}{2}ight)^2 + \left(-\frac{1}{2}ight) + \frac{11}{12} = \frac{1}{4} - \frac{1}{2} + \frac{11}{12} = \frac{3 - 6 + 11}{12} = \frac{8}{12} = \frac{2}{3} $

Thus, the eccentricity $e$ of the ellipse is $\frac{2}{3}$, so $e^2 = \left(\frac{2}{3}ight)^2 = \frac{4}{9}$.

Using the eccentricity formula for an ellipse:

$ e^2 = \frac{1 - b^2}{a^2} \quad \Rightarrow \quad \frac{4}{9} = \frac{1 - b^2}{a^2} $

Rearranging gives:

$ b^2 = a^2 \left(1 - \frac{4}{9}ight) = a^2 \cdot \frac{5}{9} $

From equation $(1)$, $b^2 = 5a$. Substituting, we have:

$ 5a = a^2 \cdot \frac{5}{9} \quad \Rightarrow \quad a^2 = 9a $

Solving for $a$,

$ a = 9, \quad 	ext{and therefore } b^2 = 5a = 45 \quad \Rightarrow \quad b = \sqrt{45} = 3\sqrt{5} $

Finally, calculate $a^2 + b^2$:

$ a^2 + b^2 = 81 + 45 = 126 $

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