Find the equation of the hyperbola if foci are $\left(\pm 3, 5\right)$ and $e = 3$.
Given: Foci $F_1 = \left(3, 5\right)$ and $F_2 = \left(-3, 5\right)$; $\;$ $e = 3$
Distance between foci $= F_1 F_2 = \sqrt{\left(3 + 3\right)^2 + \left(5 - 5\right)^2}$
i.e. $\;$ $F_1 F_2 = 6$ $\;\;\; \cdots \; (1)$
But $F_1 F_2 = 2 a e$ $\;\;\; \cdots \; (2)$
$\therefore$ $\;$ We have from equations $(1)$ and $(2)$,
$2a \times 3 = 6$ $\implies$ $a = 1$
Now, $b^2 = a^2 \left(e^2 - 1\right) = 1 \times \left(9 - 1\right) = 8$
The foci are $\left(\pm 3, 5\right)$ $\implies$ the transverse axis is parallel to the X axis.
$\therefore$ $\;$ Let the equation of the required hyperbola be:
$\dfrac{\left(x - h\right)^2}{a^2} - \dfrac{\left(y - k\right)^2}{b^2} = 1$ $\;\;\; \cdots \; (3)$
Center $C \left(h, k\right)$ is the midpoint of $F_1$ and $F_2$.
$\therefore$ $\;$ $C\left(h, k\right) = \left(\dfrac{3 - 3}{2}, \dfrac{5 + 5}{2}\right) = \left(0, 5\right)$
$\therefore$ $\;$ Equation of the required hyperbola is
$\dfrac{\left(x - 0\right)^2}{1^2} - \dfrac{\left(y - 5\right)^2}{8} = 1$
i.e. $\;$ $\dfrac{x^2}{1} - \dfrac{\left(y - 5\right)^2}{8} = 1$