Let a hyperbola in the standard form have eccentricity 3 2 and the length of its latus rectum be 5 . Let P be a point on the hyperbola in the first quadrant such that the area of the triangle formed by P and the two foci is 3 5 . Then the square of the distance of P from the origin is equal to :

Step-by-step solution

Let the equation of the hyperbola be x^2 a^2 - y^2 b^2 = 1 . Given eccentricity e = 3 2 , we have b^2 = a^2(e^2 - 1) = a^2 ( 9 4 - 1 ) = 5a^2 4 . The length of the latus rectum is 2b^2 a = 5 . Substituting b^2 , we get 2 a ( 5a^2 4 ) = 5 5a 2 = 5 a = 2 . Then b^2 = 5(4) 4 = 5 . The distance of the focus from the center is c = ae = 2 ( 3 2 ) = 3 . The foci are S(3, 0) and S'(-3, 0) , so the base of the triangle SS' = 6 . Let the coordinates of P be (x₁, y₁) . The area of PSS' is 1 2 SS' y₁ = 3 5 . 1 2 6 y₁ = 3 5 y₁