The hyperbolas $H_1: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and $H_2: \frac{x^2}{a^2} - \frac{y^2}{b^2} = -1$ are shown in the diagram - HSC - SSCE Mathematics Extension 2 - Question 13 - 2015 - Paper 1

The coordinates of point $P$ on the hyperbola are $(sec\theta, b tan\theta)$ and $Q$ has coordinates $(a tan\theta, b sec\theta)$.

To find the slope of line $PQ$, we first calculate: $m = \frac{b sec\theta - b tan\theta}{a tan\theta - sec\theta} = b \frac{sec\theta - tan\theta}{a tan\theta - sec\theta}$

Using the point-slope formula, the equation of line $PQ$ can be written as: $y - b tan\theta = m(x - sec\theta)$

Expanding this gives: $y = b tan\theta + m(x - sec\theta)$

Substituting and rearranging leads to the desired form: $bx + ay = ab(tan\theta + sec\theta)$.

To find the area of triangle $OPQ$, we can use the formula: $Area = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$

Assigning coordinates: $O(0,0)$, $P(sec\theta, b tan\theta)$, and $Q(a tan\theta, b sec\theta)$, we have:

Calculating: $Area = \frac{1}{2} |0(b tan\theta - b sec\theta) + sec\theta(b sec\theta - 0) + a tan\theta(0 - b tan\theta)|$ $= \frac{1}{2} |sec\theta(b sec\theta) - a tan\theta(b tan\theta)|$ $= \frac{1}{2} |b sec^2\theta - ab tan^2\theta|$

Since $sec^2\theta - tan^2\theta = 1$, this simplifies to: $Area = \frac{1}{2} |b|$

Hence, the area is constant and does not depend on $\theta$.