Let O be the origin, and P and Q be two points on the rectangular hyperbola xy=12 such tha…

Question

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 $\left(\dfrac12,-\dfrac12\right)$. Then the area of the triangle $OPQ$ equals:

Accepted Answer

Let $P=(x_1,y_1),\ Q=(x_2,y_2)$ with $x_1y_1=x_2y_2=12$, $x_1+x_2=1,\ y_1+y_2=-1$.

$$y_1+y_2=\frac{12(x_1+x_2)}{x_1x_2}=-1 \;\Rightarrow\; \frac{12}{x_1x_2}=-1 \;\Rightarrow\; x_1x_2=-12.$$

So $x_1,x_2$ are roots of $t^2-t-12=0 \Rightarrow t=4,-3$. Taking $x_1=4,x_2=-3$: $y_1=3,\ y_2=-4$.

$$P=(4,3),\ Q=(-3,-4),\quad \text{Area}=\frac12|x_1y_2-x_2y_1|=\frac12|-16-(-9)|=\frac12(7)=\frac72.$$

Solutions · 1