Given ellipse

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\qquad a>b$

For this ellipse:

• Eccentricity: $e=\sqrt{1-\frac{b^2}{a^2}}$
• Length of latus rectum: $\frac{2b^2}{a}$

Find the eccentricity from the function

The function is $\phi(t)=\frac{5}{12}+t-t^2$

Since this is a downward opening parabola, its maximum occurs at $t=-\frac{B}{2A}=-\frac{1}{2(-1)}=\frac12$

Now, $\phi\left(\frac12\right)=\frac{5}{12}+\frac12-\frac14$

Convert to common denominator: $\frac{5}{12}+\frac{6}{12}-\frac{3}{12}=\frac{8}{12}=\frac23$

So the eccentricity is $e=\frac23$

Use eccentricity relation

We know $e^2=1-\frac{b^2}{a^2}$

Hence, $\frac{4}{9}=1-\frac{b^2}{a^2}$ $\frac{b^2}{a^2}=1-\frac49=\frac59$

Therefore, $b^2=\frac59 a^2$

Use latus rectum length

Given length of latus rectum is $10$: $\frac{2b^2}{a}=10$ $b^2=5a$

But from above, $b^2=\frac59 a^2$

So, $\frac59 a^2=5a$

Since $a>0$, $\frac59 a=5$ $a=9$

Find $b^2$

From $b^2=5a$, $b^2=5\cdot 9=45$

Also, $a^2=81$

Compute $a^2+b^2$

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

Option check

The correct option is: $\boxed{126}$ which is Option B.