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

The standard equation of an ellipse with its major axis along the $$x$$-axis is $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$, where $$a \gt b \gt 0$$ and its eccentricity is $$e$$.

1. Length of the latus-rectum (the chord through a focus perpendicular to the major axis) is
$$\text{Latus rectum}= \frac{2b^{2}}{a}$$.

2. The eccentricity and the semi-axes satisfy the relation
$$b^{2}=a^{2}(1-e^{2})$$ $$-(1)$$.

3. We are given that the length of the latus-rectum equals $$10$$:
$$\frac{2b^{2}}{a}=10$$ $$-(2)$$.

4. The eccentricity $$e$$ is the minimum value of the quadratic function
$$f(t)=t^{2}+t+\frac{11}{12},\; t\in\mathbb{R}$$.

The minimum of a quadratic $$At^{2}+Bt+C$$ with $$A\gt 0$$ occurs at $$t=-\frac{B}{2A}$$.
Here $$A=1,\; B=1,\; C=\frac{11}{12}$$, so the minimising value of $$t$$ is
$$t=-\frac{1}{2}$$.

Substituting back gives the minimum value (and hence the eccentricity):
$$e=f\!\left(-\frac{1}{2}\right) =\left(-\frac{1}{2}\right)^{2} +\left(-\frac{1}{2}\right) +\frac{11}{12} =\frac{1}{4}-\frac{1}{2}+\frac{11}{12}$$
$$=\frac{1}{4}-\frac{2}{4}+\frac{11}{12} =-\frac{1}{4}+\frac{11}{12} =\frac{-3}{12}+\frac{11}{12} =\frac{8}{12} =\frac{2}{3}$$.

Thus $$e=\frac{2}{3}$$ and $$e^{2}=\frac{4}{9}$$.

5. Using equation $$(1)$$:
$$b^{2}=a^{2}\!\left(1-\frac{4}{9}\right)=a^{2}\!\left(\frac{5}{9}\right) =\frac{5a^{2}}{9}$$ $$-(3)$$.

6. Insert $$(3)$$ into the latus-rectum condition $$(2)$$:
$$\frac{2}{a}\left(\frac{5a^{2}}{9}\right)=10 \;\;\Longrightarrow\;\; \frac{10a}{9}=10 \;\;\Longrightarrow\;\; a=\frac{10\times9}{10}=9$$.

Therefore $$a^{2}=9^{2}=81$$.

7. Find $$b^{2}$$ from $$(3)$$:
$$b^{2}=\frac{5a^{2}}{9} =\frac{5\times81}{9}=5\times9=45$$.

8. Finally,
$$a^{2}+b^{2}=81+45=126$$.

Hence $$a^{2}+b^{2}=126$$. The correct option is Option B.