Find the equation of the hyperbola if its asymptotes are parallel to $\;$ $x + 2y - 12 = 0$ $\;$ and $\;$ $x - 2y+ 8 = 0$, $\;$ $\left(2,4\right)$ is the center of the hyperbola and it passes through $\left(2,0\right)$.
Equations of the given straight lines are
$x + 2y - 12 = 0$ $\;\;\; \cdots \; (1)$ $\;$ and $\;$ $x - 2y + 8 = 0$ $\;\;\; \cdots \; (2)$
Slope of equation $(1)$ is $= m_1 = \dfrac{-1}{2}$
Slope of equation $(2)$ is $= m_2 = \dfrac{1}{2}$
$\because$ $\;$ the asymptotes are parallel to the given lines [equations $(1)$ and $(2)$],
$\therefore$ $\;$ the slopes of the asymptotes are $\dfrac{1}{2}$ and $\dfrac{-1}{2}$
Let the equations of the asymptotes be
$y = \dfrac{1}{2}x + p$ $\;\;\; \cdots \; (3a)$ $\;$ and $\;$ $y = -\dfrac{1}{2}x + q$ $\;\;\; \cdots \; (3b)$
where $p$ and $q$ are the Y intercepts.
Solving equations $(3a)$ and $(3b)$ simultaneously gives the point of intersection of the asymptotes.
$\therefore$ $\;$ We have from equations $(3a)$ and $(3b)$
$\dfrac{1}{2}x + p = \dfrac{-1}{2}x + q$ $\implies$ $x = q - p$
Substituting the value of $x$ in equation $(3a)$ gives
$y = \dfrac{1}{2} \left(q - p\right) + p$ $\implies$ $y = \dfrac{1}{2} \left(p + q\right)$
The asymptotes intersect at the center.
Given: Center of hyperbola $= \left(2, 4\right)$
$\therefore$ $\;$ We have $\;$ $q - p = 2$ $\;\;\; \cdots \; (4a)$
and $\;$ $\dfrac{1}{2} \left(p + q\right) = 4$ $\implies$ $p + q = 8$ $\;\;\; \cdots \; (4b)$
Solving equations $(4a)$ and $(4b)$ simultaneously we have,
$2q = 10$ $\implies$ $q = 5$
Substituting the value of $q$ in equation $(4a)$ gives
$p = q - 2 = 3$
$\therefore$ $\;$ The equations of the asymptotes are
$y = \dfrac{1}{2}x + 3$ $\;\;$ i.e. $\;$ $x - 2y + 6 = 0$
and $\;$ $y = \dfrac{-1}{2}x + 5$ $\;\;$ i.e. $\;$ $x + 2y - 10 = 0$
$\therefore$ $\;$ The combined equations of the asymptotes is
$\left(x - 2y + 6\right) \left(x + 2y - 10\right) = 0$ $\;\;\; \cdots \; (5)$
The equation of the hyperbola differs from the combined equation of the asymptotes by a constant.
$\therefore$ $\;$ The equation of the hyperbola is of the form
$\left(x - 2y + 6\right) \left(x + 2y - 10\right) + k = 0$ $\;\;\; \cdots \; (6)$
The required hyperbola passes through the point $\left(2, 0\right)$.
$\therefore$ $\;$ We have from equation $(6)$,
$\left(2 - 0 + 6\right) \left(2 + 0 - 10\right) + k = 0$
$\implies$ $k = 64$
Substituting the value of $k$ in equation $(6)$, the equation of the required hyperbola is
$\left(x - 2y + 6\right) \left(x + 2y - 10\right) + 64 = 0$