Question

How do you find the vertices, asymptote, foci and graph `x^2/36-y^2=1`?

Answer 1

The General form for a hyperbola of this type is:`(x-h)^2/a^2-(y-k)^2/b^2=1" [1]"`
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`

Explanation:

Here is a reference for Hyperbola equations provided in the answer section.

Let's convert the given equation, `x^2/36-y^2=1`, to the form of equation [1]:

`(x-0)^2/6^2-(y-0)^2/1^2=1" [2]"`

By inspection of equation [2] we observe that `h = k = 0, a = 6, and b = 1`

Vertices: `(-6,0)` and `(6,0)`
Foci: `(-sqrt(37),0)` and `(sqrt(37),0)`

Asymptotes: `y=1/6x and y = -1/6x`

Here is a graph:

graph{x^2/36-y^2=1 [-10, 10, -5, 5]}