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 the areas are equal, consider the definite integrals representing the areas:

1. The area of region OPQ can be expressed as:

$A_{OPQ} = \int_{q}^{p} \frac{1}{x} \, dx = \ln(p) - \ln(q) = \ln\left( \frac{p}{q} \right).$

2. The area of region Q'QP' can be calculated similarly as:

A_{Q'QP'} = \int_{q}^{p} rac{1}{x} \, dx = \ln(p) - \ln(q) = \ln\left( \frac{p}{q} \right).

Thus, since both integrals yield the same result, the areas are equal.

Consider triangle OPM and OQR where M is the midpoint of PR. Using the similarity of triangles, the relationship between the segments can be derived:

From triangle OPM, we find that:

$\frac{OM}{OP} = \frac{OR}{OQ} \implies \frac{r}{p} = \frac{OM}{OQ}$

and thus:

• Squaring both sides gives us:

$$r^2 = pq.$

To show the line RR' divides the shaded region, we compare the area of segments on either side of RR'. Setting up the integral for the area above the line RR' gives:

$A_1 = \int_{q}^{c} \frac{1}{x} \, dx\quad \text{and} \quad A_2 = \int_{c}^{p} \frac{1}{x} \, dx$

where c is the x-coordinate of point R. Setting these equal with respect to the total area will yield:

( A_1 = A_2 ) showing that RR' splits the area equally.

Using the results from previous parts, particularly the equal division of area by RR', we can conclude that because the areas OPQ and Q'QP' are equal and RR' acts as a symmetry line for the shaded regions, the line OR must also divide the region OPQ into two equal areas by symmetry arguments.