Find the equation of the hyperbola if the center is $\left(1, 4\right)$; one of the foci is $\left(6, 4\right)$ and the corresponding directrix is $x = \dfrac{9}{4}$
Given: Directrix of hyperbola is: $\;$ $x = \dfrac{9}{4}$
$\implies$ the directrix is parallel to the Y axis.
$\therefore$ $\;$ Let the equation of the required hyperbola be:
$\dfrac{\left(x - h\right)^2}{a^2} - \dfrac{\left(y - k\right)^2}{b^2} = 1$ $\;\;\; \cdots \; (1)$
Given: Center $C \left(h, k\right) = \left(1, 4\right)$ $\;\;\; \cdots \; (2a)$; Focus $F_1 = \left(6, 4\right)$ $\;\;\; \cdots \; (2b)$
$CF_1 = a \cdot e = \sqrt{\left(6 - 1\right)^2 + \left(4 - 4\right)^2}$
i.e. $\;$ $a \cdot e = 5$ $\;\;\; \cdots \; (3)$
Draw CZ perpendicular to the directrix.
Then, $Z = \left(\dfrac{9}{4}, 4\right)$
Distance between center and directrix is $= CZ = \dfrac{a}{e} = \sqrt{\left(\dfrac{9}{4} - 1\right)^2 + \left(4 - 4\right)^2}$
i.e. $\;$ $\dfrac{a}{e} = \dfrac{5}{4}$ $\;\;\; \cdots \; (4)$
$\therefore$ $\;$ We have from equations $(3)$ and $(4)$,
$a \cdot e \times \dfrac{a}{e} = 5 \times \dfrac{5}{4}$
i.e. $\;$ $a^2 = \dfrac{25}{4}$ $\;\;\; \cdots \; (5a)$
and $\dfrac{a \cdot e}{a / e} = \dfrac{5}{5 / 4}$ $\implies$ $e^2 = 4$
Now, $b^2 = a^2 \times \left(e^2 - 1\right) = \dfrac{25}{4} \times \left(4 - 1\right)$
i.e. $\;$ $b^2 = \dfrac{75}{4}$ $\;\;\; \cdots \; (5b)$
$\therefore$ $\;$ In view of equations $(2a)$, $(5a)$ and $(5b)$, the equation of required hyperbola [equation $(1)$] becomes
$\dfrac{\left(x -1 \right)^2}{25 / 4} - \dfrac{\left(y - 4\right)^2}{75 / 4} = 1$