Let S and S^ be the foci of the ellipse x^ 2 25 + y^ 2 9 =1 and...

For the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$, we have $a = 5$, $b = 3$, $c = 4$.
So foci are at $S(\pm 4, 0)$.
For point $P$ on the ellipse: $SP + S'P = 10$.
Given $(SP)^2 + (S'P)^2 - SP \cdot S'P = 37$
Let $r_1 = SP$ and $r_2 = S'P$.
From $(r_1 + r_2)^2 = 100$, we get $r_1^2 + r_2^2 = 100 - 2r_1r_2$.
Substituting into the given equation: $100 - 3r_1r_2 = 37$, so $r_1r_2 = 21$.
Thus $r_1 = 3, r_2 = 7$.
From $(\alpha + 4)^2 + \beta^2 = 9$ and $(\alpha - 4)^2 + \beta^2 = 49$, subtracting gives $16\alpha = -40$, so $\alpha = -\frac{5}{2}$.
From the ellipse equation: $\beta^2 = \frac{27}{4}$.
Therefore $\alpha^2 + \beta^2 = \frac{25}{4} + \frac{27}{4} = 13$.