Let an ellipse be given by x^2 a^2 + y^2 16 = 1 , where a > 4 . A triangle is inscribed in the ellipse with one of its vertices at A(a, 0) and the opposite side PQ parallel to the y -axis. If the maximum possible area of this triangle is 15 3 , and the tangents to the ellipse at P and Q intersect at a point R , then the distance between R and the focus of the ellipse having a negative x -coordinate is

Step-by-step solution

Let P(a , 4 ) and Q(a , -4 ) . The area of APQ is A = 1 2 (8 ) (a - a ) = 4a (1 - ) . For maximum area, dA d = 4a( - 2 ) = 0 = - 1 2 = 2 3 . Maximum area = 4a ( 3 2 ) (1 - (- 1 2 ) ) = 3 3 a . Given 3 3 a = 15 3 a = 5 . The equation of the ellipse is x^2 25 + y^2 16 = 1 . The line PQ is the chord of contact for the point R . Since the x -coordinate of P and Q is 5 (- 1 2 ) = - 5 2 , the equation of PQ is x = - 5 2 . Let R be (x₁, y₁) . The chord of contact is xx₁ 25 + yy₁ 16 = 1 . Comparing with x = - 5 2 , we get