Let S and S' be the foci of the ellipse $$\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$$ and $$P(\alpha , \beta)$$ be a point on the ellipse in the first quadrant. If $$(SP)^{2}+(S'P)^{2}-SP\cdot S'P=37$$, then $$\alpha^{2}+\beta^{2}$$ is equal to :

We need to determine the value of $$\alpha^2 + \beta^2$$ for a point $$P(\alpha, \beta)$$ on the ellipse $$\frac{x^2}{25} + \frac{y^2}{9} = 1$$ satisfying a specific condition on its distances from the foci.

For this ellipse one has $$a^2 = 25, b^2 = 9$$ so that $$a = 5, b = 3$$, and the focal distance is $$c = \sqrt{a^2 - b^2} = \sqrt{25 - 9} = 4$$. Hence the foci lie at $$S(4,0)$$ and $$S'(-4,0)$$, and by definition of the ellipse, any point $$P$$ on it satisfies $$SP + S'P = 2a = 10$$.

Denoting $$s = SP$$ and $$s' = S'P$$, the problem states that $$s^2 + s'^2 - s s' = 37$$ in addition to $$s + s' = 10$$.

Expanding $$(s + s')^2 = s^2 + 2ss' + s'^2 = 100$$ gives $$s^2 + s'^2 = 100 - 2ss'$$. Substituting into the condition $$s^2 + s'^2 - ss' = 37$$ yields $$(100 - 2ss') - ss' = 37$$, so $$100 - 3ss' = 37$$ and hence $$ss' = 21$$.

On the other hand, expressing the squares of the focal distances in terms of the coordinates of $$P$$ gives $$SP^2 = (\alpha - 4)^2 + \beta^2 = \alpha^2 - 8\alpha + 16 + \beta^2$$ and $$S'P^2 = (\alpha + 4)^2 + \beta^2 = \alpha^2 + 8\alpha + 16 + \beta^2$$. Adding these results in $$SP^2 + S'P^2 = 2\alpha^2 + 2\beta^2 + 32$$, while also $$SP^2 + S'P^2 = (s + s')^2 - 2ss' = 100 - 42 = 58$$. Equating gives $$2\alpha^2 + 2\beta^2 + 32 = 58$$, so $$2(\alpha^2 + \beta^2) = 26$$ and therefore $$\alpha^2 + \beta^2 = 13$$.

Thus the required value is 13.