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

1 Answer
Douglas K.
Dec 6, 2016

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:
Desmos.com

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