Question

How do you write an equation for a hyperbola with vertices (1, 3) and (-5, 3), and foci (3, 3) and (-7, 3)?

Answer 1

The equation is:

`(x - -2)^2/3^2 - (y - 3)^2/4^2 = 1`

Explanation:

Please notice that the vertices are of the forms:

`(h - a, k)` and `(h + a,k)` specifically `(-5,3)` and `(1, 3)`

The same information can be deduced from the foci, which have the forms:

`(h - c, k)` and `(h + c, k)` specifically `(-7, 3)` and `(3, 3)`

The standard form for the equation of a hyperbola, where the vertices and foci have these properties, is the horizontal transverse axis form:

`(x - h)^2/a^2 - (y -k)^2/b^2 = 1`

`k = 3` by observation:

`(x - h)^2/a^2 - (y - 3)^2/b^2 = 1`

Compute h and a:

`-5 = h - a` and `1 = h + a`

`2h = -4`

`h = -2`

`a = 3`

`(x - -2)^2/3^2 - (y - 3)^2/b^2 = 1`

To complete the equation, we only need the value of b but, to find the value of b, we must, first, find the value of c:

Using the `(h + c, k)` form for the focus point, `(3,3)`, we substitute -2 for h, set the right side equal to 3, and then solve for c:

`-2 + c = 3`

`c = 5`

Solve for b, using the equation `c^2 = a^2 + b^2`:

`5^2 = 3^2 + b^2`

`b = 4`

`(x - -2)^2/3^2 - (y - 3)^2/4^2 = 1`