Let P (10,2 15 ) be a point on the hyperbola x² a ² - y² ~b ² =1 , whose foci are S and S ^ . If the length of its latus rectum is 8, then the square of the area of PSS ^ is equal to :

Step-by-step solution

For the hyperbola x^2 a^2 - y^2 b^2 = 1 , point P(10, 2 15 ) gives 100 a^2 - 60 b^2 = 1 . The latus rectum length 2b^2 a = 8 gives b^2 = 4a . Substituting: 100 a^2 - 15 a = 1 , which yields a^2 + 15a - 100 = 0 , so a = 5 and b^2 = 20 . Thus c^2 = 45 and c = 3 5 . The foci are S( 3 5 , 0) . Triangle PSS' has base SS' = 6 5 and height 2 15 . Area = 1 2 6 5 2 15 = 30 3 . Therefore (Area) ^2 = 2700 .