Let S and S' be the foci of an ellipse and P be a point on it. The maximum area of the triangle PSS' is 12 and the maximum value of the product SP S'P is 25 . If the eccentricity e of the ellipse satisfies e > 1 2 , then the minimum value of the product SP S'P is equal to :

Step-by-step solution

Let the ellipse be x^2 a^2 + y^2 b^2 = 1 ( a > b ). The area of the triangle PSS' is 1 2 ( base ) ( height ) = 1 2 2ae |y| = ae|y| . The maximum area occurs when |y| is maximum, which is at the co-vertices where |y| = b . Thus, the maximum area is abe . Given abe = 12 . The product of focal distances for a point P(x, y) is SP S'P = a^2 - e^2x^2 . Its maximum value is a^2 (at x = 0 ). Given a^2 = 25 a = 5 . Substituting a = 5 into abe = 12 , we get 5be = 12 be = 12 5 . Squaring both sides, b^2e^2 = 144 25 . Using th