Let $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ $$(a > b)$$ be a given ellipse, length of whose latus rectum is 10. If its eccentricity is the maximum value of the function, $$\phi(t) = \frac{5}{12} + t - t^2$$, then $$a^2 + b^2$$ is equal to:

We have an ellipse whose standard equation is $$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$$ with the major axis along the x-direction and the condition $$a>b\;.$$

The question gives two independent pieces of information:

1. The length of the latus-rectum is 10.
2. The eccentricity is the greatest value attained by the quadratic function $$\phi(t)=\dfrac{5}{12}+t-t^{2}\,.$$

We shall translate each of these statements into equations involving $$a,\;b$$ and then solve for $$a^{2}+b^{2}.$$

Length of the latus-rectum.
For an ellipse centred at the origin, the length of the latus-rectum (the complete chord through a focus perpendicular to the major axis) is given by the well-known formula

$$\text{Length of latus-rectum}= \dfrac{2b^{2}}{a}\,.$$

The problem states that this length equals 10, so we write

$$\dfrac{2b^{2}}{a}=10 \;\Longrightarrow\; b^{2}=5a.$$

We shall remember this linear relation between $$b^{2}$$ and $$a$$ for later substitution.

Maximum value of the given quadratic and the eccentricity.
The eccentricity of the ellipse, denoted $$e,$$ is declared to be the maximum value of $$\phi(t)=\dfrac{5}{12}+t-t^{2}.$$

First we recall the standard result for a quadratic $$f(t)=At^{2}+Bt+C$$ with $$A\lt 0:$$ the maximum occurs at

$$t=-\dfrac{B}{2A}$$

and the corresponding maximum value is obtained by substituting this $$t$$ back into the quadratic.

In the present case we have $$A=-1,\;B=1,\;C=\dfrac{5}{12}.$$ Hence

$$t_{\text{max}}=-\dfrac{B}{2A}=-\dfrac{1}{2(-1)}=\dfrac12.$$

Substituting $$t=\dfrac12$$ into $$\phi(t)$$ gives

$$\phi_{\text{max}}=\dfrac{5}{12}+\dfrac12-\left(\dfrac12\right)^{2} =\dfrac{5}{12}+\dfrac12-\dfrac14 =\dfrac{5}{12}+\dfrac{6}{12}-\dfrac{3}{12} =\dfrac{8}{12} =\dfrac23.$$

Therefore the eccentricity is

$$e=\dfrac23.$$

Relating $$a$$ and $$b$$ using the eccentricity.
For an ellipse with semimajor axis $$a$$ and semiminor axis $$b,$$ the eccentricity is defined by

$$e=\sqrt{1-\dfrac{b^{2}}{a^{2}}}\,.$$

Squaring both sides, we have

$$e^{2}=1-\dfrac{b^{2}}{a^{2}} \;\Longrightarrow\; \dfrac{b^{2}}{a^{2}}=1-e^{2}.$$

Since $$e=\dfrac23,$$ we compute

$$e^{2}=\left(\dfrac23\right)^{2}=\dfrac49,$$ so

$$\dfrac{b^{2}}{a^{2}}=1-\dfrac49=\dfrac59.$$

This yields the quadratic relation

$$b^{2}=\dfrac59\,a^{2}.$$

Solving simultaneously for $$a$$ and $$b.$$
We have already obtained a linear relation $$b^{2}=5a$$ from the latus-rectum condition. Setting this equal to the quadratic relation just found, we write

$$5a=\dfrac59\,a^{2}.$$

Dividing both sides by 5 gives

$$a=\dfrac19\,a^{2}.$$

Because $$a\neq0,$$ we may divide by $$a$$ to obtain

$$1=\dfrac19\,a \;\Longrightarrow\; a=9.$$

With $$a$$ now known, we return to $$b^{2}=5a$$ to find $$b^{2}:$$

$$b^{2}=5a=5\times9=45.$$

Computing $$a^{2}+b^{2}.$$
We finally evaluate

$$a^{2}+b^{2}=9^{2}+45=81+45=126.$$

Hence, the correct answer is Option C.