Let P (10,2 15 ) be a point on the hyperbola x^ 2 a ^ 2 - y^ 2 ~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 :

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Quantrex Academy · Accepted Answer

Correct answer: D. 2700

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.

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