Let P(10, 2 15) be a point on the hyperbola x2a2 - y2b2 = 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 :
- 4200
- 1462
- 900
- 2700
Answer
(4) 2700
Solution
Latus rectum: 2b2a = 8 b2 = 4a
P lies on the hyperbola: 100a2 - 604a = 1
100 - 15a = a2 a2 + 15a - 100 = 0
(a + 20)(a - 5) = 0 a = 5 (as a > 0), b2 = 20
a2e2 = a2 + b2 = 25 + 20 = 45 ae = 3 5
SS' = 2ae = 6 5 (along the x-axis)
Height of P above SS′ = 2 15
Area of PSS' = 12 6 5 2 15 = 6 75 = 30 3
(Area)2 = 900 3 = 2700
Answer: (4)