# How do you find an equation of the hyperbola given Vertices: (2,3), (2,-3); Foci: (2,5), (2,-5)?

First note that the focal axis for this hyperbola is the vertical line $x = 2$ and that the "center" of the hyperbola is the point $\left(x , y\right) = \left(2 , 0\right)$.
The equation of the hyperbola therefore takes the form $- {\left(\frac{x - 2}{a}\right)}^{2} + {\left(\frac{y}{b}\right)}^{2} = 1$. The fact that the vertices are at $\left(2 , \setminus \pm 3\right)$ means that ${0}^{2} + {\left(\frac{3}{b}\right)}^{2} = 1$ so that we can take $b = 3$.
The foci being at the points $\left(2 , \setminus \pm 5\right)$ implies that $\sqrt{{a}^{2} + {b}^{2}} = 5$ so that ${a}^{2} + 9 = 25$, ${a}^{2} = 16$, and we can take $a = 4$.
The final answer can be then written as $- {\left(\frac{x - 2}{4}\right)}^{2} + {\left(\frac{y}{3}\right)}^{2} = 1$. This can also be written as $- 9 {\left(x - 2\right)}^{2} + 16 {y}^{2} = 144$ and as $- 9 {x}^{2} + 36 x - 36 + 16 {y}^{2} = 144$ and also as $9 {x}^{2} - 36 x - 16 {y}^{2} + 180 = 0$.