Let O be the origin, and P and Q be two points on the rectangular hyperbola xy = 12 such that the mid point of the line segment PQ is ( 1 2 , - 1 2 ) . Then the area of the triangle OPQ equals:

Step-by-step solution

The equation of the chord of the hyperbola xy = 12 with midpoint (x₁, y₁) is given by T = S₁ . x y₁ + y x₁ 2 - 12 = x₁ y₁ - 12 x y₁ + y x₁ = 2 x₁ y₁ Substituting the midpoint ( 1 2 , - 1 2 ) : x (- 1 2 ) + y ( 1 2 ) = 2 ( 1 2 ) (- 1 2 ) -x + y = -1 y = x - 1 To find the coordinates of P and Q, substitute y = x - 1 into the equation of the hyperbola xy = 12 : x(x - 1) = 12 x^2 - x - 12 = 0 (x - 4)(x + 3) = 0 x = 4, -3 For x = 4 , y = 3 . For x = -3 , y = -4 . Thus, the coordinates of P and Q are (4, 3) and (-3, -4)