Area of Triangle Formed by Origin and Hyperbola Segment

To find the area of the triangle $OPQ$, we first determine the equation of the line segment $PQ$ passing through the given midpoint.

The given equation of the rectangular hyperbola is: $xy = 12 \implies xy - 12 = 0$

Let the midpoint of the chord $PQ$ be $M(h, k) = \left(\frac{1}{2}, -\frac{1}{2}\right)$.

The equation of a chord of the hyperbola $xy = c^2$ with a given midpoint $(h, k)$ is given by the formula: $T = S_1$

For $xy - 12 = 0$, $T$ and $S_1$ are defined as: $T = \frac{x k + y h}{2} - 12$ $S_1 = h k - 12$

Equating $T = S_1$: $\frac{x k + y h}{2} - 12 = h k - 12$ $x k + y h = 2 h k$

Substituting $h = \frac{1}{2}$ and $k = -\frac{1}{2}$: $x\left(-\frac{1}{2}\right) + y\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)$ $-\frac{x}{2} + \frac{y}{2} = -\frac{1}{2}$ $x - y = 1 \implies y = x - 1$

To find the points of intersection $P$ and $Q$, we substitute $y = x - 1$ into the equation of the hyperbola $xy = 12$: $x(x - 1) = 12$ $x^2 - x - 12 = 0$

Factoring the quadratic equation: $(x - 4)(x + 3) = 0$

Thus, the $x$-coordinates of $P$ and $Q$ are $x_1 = 4$ and $x_2 = -3$.

Using $y = x - 1$:

• For $x_1 = 4$, $y_1 = 4 - 1 = 3 \implies P = (4, 3)$
• For $x_2 = -3$, $y_2 = -3 - 1 = -4 \implies Q = (-3, -4)$

The vertices of triangle $OPQ$ are $O(0, 0)$, $P(4, 3)$, and $Q(-3, -4)$.

The area of triangle $OPQ$ with origin $O(0, 0)$ is given by: $\text{Area}(\triangle OPQ) = \frac{1}{2} |x_1 y_2 - x_2 y_1|$

Substituting the coordinates: $\text{Area}(\triangle OPQ) = \frac{1}{2} |(4)(-4) - (-3)(3)|$ $\text{Area}(\triangle OPQ) = \frac{1}{2} |-16 + 9| = \frac{1}{2} |-7| = \frac{7}{2}$

Thus, the area of the triangle $OPQ$ is $\frac{7}{2}$.