Let the length of a latus rectum of an ellipse x²/a² + y²/b² = 1 be 10. If its eccentricity is the minimum value of the function f(t) = t² + t + 11/12, t ∈ R, then a² + b² is equal to:

 Question:

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

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Short Solution (Text):

Step 1 — Minimum value of $f(t)$

For $f(t)=t^2+t+\dfrac{11}{12}$, the vertex (minimum) occurs at $t=-\dfrac{1}{2}$.
Minimum value:

$$f\text{\hspace{0.17em}⁣}(-\frac{1}{2})={(-\frac{1}{2})}^2+(-\frac{1}{2})+\frac{11}{12}=\frac{1}{4}-\frac{1}{2}+\frac{11}{12}=\frac{3+(-6)+11}{12}=\frac{8}{12}=\frac{2}{3}\mathrm{.}$$

So the eccentricity $e=\dfrac{2}{3}$
Thus $e^2=\dfrac{4}{9}$

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Step 2 — Relation between $a,b$ and $e$

For the ellipse (assuming $a$ is semi-major, $a>b$):

$$e^2=1-\frac{b^2}{a^2}\Rightarrow \frac{b^2}{a^2}=1-e^2=1-\frac{4}{9}=\frac{5}{9}\mathrm{.}$$

So

$$\begin{array}{cccc}& b^2=\frac{5}{9}{a}^2\mathrm{.}& & \text{(1)}\end{array}$$

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Step 3 — Use latus rectum length

Latus rectum length for this ellipse = $\dfrac{2b^2}{a}=10$
Hence

$$\begin{array}{cccc}& b^2=5a\mathrm{.}& & \text{(2)}\end{array}$$

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Step 4 — Solve (1) & (2)

From (1) and (2):

$$\frac{5}{9}{a}^2=5a\Rightarrow \frac{1}{9}{a}^2=a\Rightarrow a\mathrm{\ne }0\Rightarrow a=9.$$

Then $a^2=81$. From (2): $b^2=5a=5\times9=45$.

Therefore

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

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✅ Final Answer: $\boxed{126}$

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