Question

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

Answer 1

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 `(x,y)=(2,0)`.

The equation of the hyperbola therefore takes the form `-((x-2)/a)^2+(y/b)^2=1`. The fact that the vertices are at `(2,\pm 3)` means that `0^2+(3/b)^2=1` so that we can take `b=3`.

The foci being at the points `(2,\pm 5)` 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 `-((x-2)/4)^2+(y/3)^2=1`. This can also be written as `-9(x-2)^2+16y^2=144` and as `-9x^2+36x-36+16y^2=144` and also as `9x^2-36x-16y^2+180=0`.