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:

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Correct answer: C. 7 2

The equation of the chord of the hyperbola xy = 12 with midpoint (x_1, y_1) is given by T = S_1. x y_1 + y x_1 2 - 12 = x_1 y_1 - 12 x y_1 + y x_1 = 2 x_1 y_1 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). The area of triangle OPQ with vertices (0,0), (4,3), and (-3,-4) is: Area = 1 2 |x_P y_Q - x_Q y_P| Area = 1 2 |(4)(-4) - (-3)(3)| Area = 1 2 |-16 + 9| = 7 2 Answer: 7 2

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