Find the equation of the rectangular hyperbola which has its center at $\left(2,1\right)$, one of its asymptotes $3x - y - 5 = 0$ and which passes through the point $\left(1, -1\right)$.
Let the equation of the required rectangular hyperbola be
$\left(x - h\right) \left(y - k\right) = c^2$ $\;\;\; \cdots \; (1)$
One of the asymptotes of the required rectangular hyperbola is
$3x - y - 5 = 0$ $\;\;\; \cdots \; (2)$
Slope of the given asymptote is $= m_1 = 3$
The asymptotes of a rectangular hyperbola are perpendicular to each other.
$\therefore$ $\;$ Slope of the second asymptote $= m_2 = \dfrac{-1}{m_1} = \dfrac{-1}{3}$
Let the equation of the second asymptote be
$y = \dfrac{-1}{3}x + p$ $\;$ where $p$ is a constant.
i.e. $\;$ $x + 3y = 3p$ $\;\;\; \cdots \; (3)$
Solving equations $(2)$ and $(3)$ simultaneously we have
$x = \dfrac{15 + 3p}{10}$
The asymptotes intersect at the center.
Given: Center $= \left(h, k\right) = \left(2, 1\right)$
$\implies$ $\dfrac{15 + 3p}{10} = 2$ $\implies$ $p = \dfrac{5}{3}$
Substituting the value of $p$ in equation $(3)$, the equation of the second asymptote is
$x + 3y - 5 = 0$ $\;\;\; \cdots \; (4)$
$\therefore$ $\;$ The combined equation of the asymptotes is
$\left(3x - y - 5\right) \left(x + 3y - 5\right) = 0$ $\;\;\; \cdots \; (5)$
The equation of rectangular hyperbola differs from the combined equation of asymptotes by a constant.
$\therefore$ $\;$ The equation of the rectangular hyperbola is
$\left(3x - y - 5\right) \left(x + 3y - 5\right) + \ell = 0$ $\;\;\; \cdots \; (6)$
Equation $(6)$ passes through the point $\left(1, -1\right)$.
$\therefore$ $\;$ We have
$\left(3 + 1 - 5\right) \left(1 - 3 - 5\right) + \ell = 0$ $\implies$ $\ell = -7$
$\therefore$ $\;$ The required equation of rectangular hyperbola is
$\left(3x - y - 5\right) \left(x + 3y - 5\right) - 7 = 0$