Question

How do you find the vertices, asymptote, foci and graph `x^2/64-(9y^2)/4=1`?

Answer 1

Please see the explanation.

Explanation:

Standard form for a Hyperbola of this type (Horizontal Transverse Axis) is:

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

The center is at: `(h, k)`
The vertices are at: `(h - a, k) and (h + a, k)`
The foci are at: `(h - sqrt(a^2 + b^2), k) and (h + sqrt(a^2 + b^2), k)`
The asymptotes are: `y = -b/a(x - h) + k and y = b/a(x - h) + k`

Let put the given equation in standard form:

`(x - 0)^2/8^2 - (y - 0)^2/2^2 = 1`

The center is at: `(0, 0)`
The vertices are at: `(-8, 0) and (8, 0)`
The foci are at: `(-sqrt(68), 0) and (sqrt(68), 0)`
The asymptotes are: `y = -1/4x and y = 1/4x`

Graph: