Let O be the origin, and P and Q be two points on the rectangular hyperbola xy=12 such that the midpoint of the line segment PQ is ( 12, - 12 ). Then the area of the triangle OPQ equals :
- 32
- 52
- 72
- 92
Answer
(3) 72
Solution
Let P (p, 12p ) and Q (q, 12q ).
Midpoint: p+q2= 12 p+q=1
12 ( 12p+ 12q )=- 12 12(p+q)pq=-1 pq=-12
p, q are roots of t2-t-12=0 p=4, q=-3
P(4, 3), Q(-3, -4)
Area of OPQ = 12|x1y2-x2y1|= 12|4(-4)-(-3)(3)|= 12|-16+9|= 72
Answer: (3)