Let PQ be a chord of the hyperbola x24 - y2b2 = 1, perpendicular to the x-axis such that OPQ is an equilateral triangle, O being the centre of the hyperbola. If the eccentricity of the hyperbola is 3, then the area of the triangle OPQ is
- 2 3
- 115
- 8 35
- 95
Answer
(3) 8 35
Solution
e2 = 1 + b24 = 3 b2 = 8
Let P(h, k), Q(h, -k) (k > 0), so PQ = 2k.
Equilateral: OP = PQ
h2 + k2 = 4k2 h2 = 3k2
P lies on the hyperbola: 3k24 - k28 = 1
5k28 = 1 k2 = 85
Area = 34(2k)2 = 3k2 = 8 35
Answer: (3)