JEE MainMathematicsHyperbola
Let P be a point in the first quadrant on the hyperbola 9x^2 - 16y^2 = 144 . If the area of the triangle formed by P and the two foci of the hyperbola is 15 , then the product of the focal distances of P is equal to :
Options
- A41
- B34
- C16
- D9
Correct answer
B. 34
Step-by-step solution
The equation of the hyperbola can be rewritten as x^2 16 - y^2 9 = 1 . Here, a^2 = 16 and b^2 = 9 . The distance of the foci from the center is given by c = a^2 + b^2 = 16 + 9 = 5 . The foci are S(5, 0) and S'(-5, 0) , so the distance between them is SS' = 10 . Let P(x₁, y₁) be the point on the hyperbola. The area of PSS' is given by 1 2 SS' y₁ = 15 . 1 2 10 y₁ = 15 5y₁ = 15 y₁ = 3 . Since P lies on the hyperbola, substitute y₁ = 3 into the equation: x₁^2 16 - 9 9 = 1 x₁^2 16 - 1 = 1 x₁^2 16 = 2 x₁^2 = 32 . The foc