Ellipse

Concept:Use ellipse properties: point on ellipse satisfies its equation, sum of distances from foci equals $2a$, and focal distances can be expressed in terms of $x$-coordinate.Explanation:Given ellipse: $\frac{x^2}{25}+\frac{y^2}{9}=1$, so $a=5$, $b=3$, eccentricity $e=\sqrt{1-\frac{b^2}{a^2}}=\frac{4}{5}$.$P(\alpha,\beta)$ lies on ellipse $\Rightarrow \frac{\alpha^2}{25}+\frac{\beta^2}{9}=1$.Sum of focal distances: $PS + PS' = 2a = 10$.Given: $(SP)^2+(S'P)^2 - SP\cdot S'P = 37$.Use identity: $(PS+PS')^2 = PS^2+PS'^2+2 PS\cdot PS'$ $\Rightarrow PS^2+PS'^2 = 100 - 2 PS\cdot PS'$.Substitute: $(100 - 2 PS\cdot PS') - PS\cdot PS' = 37$ $\Rightarrow 100 - 3 PS\cdot PS' = 37$ $\Rightarrow 3 PS\cdot PS' = 63$ $\Rightarrow PS\cdot PS' = 21$.For ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with foci at $(\pm ae,0)$, distances from foci for point $(\alpha,\beta)$ are: $PS = a+e\alpha$, $PS' = a-e\alpha$ (since $\alpha>0$ in first quadrant). Thus $PS = 5+\frac{4}{5}\alpha$, $PS' = 5-\frac{4}{5}\alpha$.Product: $PS\cdot PS' = \left(5+\frac{4}{5}\alpha\right)\left(5-\frac{4}{5}\alpha\right) = 25 - \frac{16}{25}\alpha^2$.Set equal to 21: $25 - \frac{16}{25}\alpha^2 = 21$ $\Rightarrow \frac{16}{25}\alpha^2 = 4$ $\Rightarrow \alpha^2 = \frac{25}{4}$.From ellipse equation: $\frac{\beta^2}{9} = 1 - \frac{\alpha^2}{25} = 1 - \frac{1}{4} = \frac{3}{4}$ $\Rightarrow \beta^2 = \frac{27}{4}$.Thus $\alpha^2+\beta^2 = \frac{25}{4}+\frac{27}{4} = \frac{52}{4} = 13$.Answer:Option A: 13.