Question

How do you find the vertices and foci of `x^2/9-y^2/64=1`?

Answer 1

Please read the Hyperbola reference and the explanation (below).

Explanation:

The general form for a hyperbola with a horizontal transverse axis is:

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

Center: `(h, k)`
Vertices: `(h - a, k) and (h + a, k)`
Foci: `(h - sqrt(a^2 + b^2), k) and (h + sqrt(a^2 + b^2), k)`
Asymptotes: `y = -b/a(x - h) + k and y = b/a(x - h) + k`

Write the given equation in the form of equation [1]:

`(x - 0)^2/3^2 - (y - 0)^2/8^2 = 1" [2]"`

Vertices: `(-3, 0) and (3, 0)`
Foci: `(-sqrt(73), 0) and (sqrt(73), 0)`

Here is a graph with the vertices and the foci.