Find the equation of the rectangular hyperbola which which has for one of its asymptotes the line $x + 2y - 5 = 0$ and passes through the points $\left(6,0\right)$ and $\left(-3,0\right)$.
Given: One of the asymptotes of the rectangular hyperbola is
$x + 2y - 5 = 0$ $\;\;\; \cdots \; (1)$
$\therefore$ $\;$ Slope of the asymptote $= m_1 = \dfrac{-1}{2}$
The asymptotes of a rectangular hyperbola are at right angles to one another.
$\therefore$ $\;$ Slope of the second asymptote $= m_2 = \dfrac{-1}{m_1} = 2$
$\therefore$ $\;$ Let the equation of the second asymptote be
$y = 2x + k$ $\;$ where $k$ is a constant.
i.e. $\;$ $2x - y + k = 0$ $\;\;\; \cdots \; (2)$
$\therefore$ $\;$ The combined equation of the asymptotes is
$\left(x + 2y - 5\right) \left(2x - y + k\right) = 0$ $\;\;\; \cdots \; (3)$
The equation of rectangular hyperbola differs from the combined equation of asymptotes by a constant.
$\therefore$ $\;$ The equation of hyperbola is
$\left(x + 2y - 5\right) \left(2x - y + k\right) + \ell = 0$ $\;\;\; \cdots \; (4)$
Equation $(4)$ passes through the points $\left(6, 0\right)$ and $\left(-3, 0\right)$.
$\therefore$ $\;$ We have
$\left(6 + 0 - 5\right) \left(12 - 0 + k\right) + \ell = 0$ $\implies$ $k + \ell = - 12$ $\;\;\; \cdots \; (5a)$
and $\left(-3 + 0 - 5\right) \left(-6 - 0 + k\right) + \ell = 0$ $\implies$ $-8k + \ell = -48$ $\;\;\; \cdots \; (5b)$
Solving equations $(5a)$ and $(5b)$ simultaneously we have,
$k = 4$ $\;$ and $\;$ $\ell = -16$
Substituting the values of $k$ and $\ell$ in equation $(4)$, the equation of the required rectangular hyperbola is
$\left(x + 2y - 5\right) \left(2x - y + 4\right) - 16 = 0$
i.e. $\left(x + 2y - 5\right) \left(2x - y + 4\right) = 16$