How do you graph #x^2/64-y^2=1# and identify the foci and asympototes?

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Answer 1 · 2017-05-27T16:50:38

For a hyperbola of the form:
#(x-h)^2/a^2-(y-k)^2/b^2=1" [1]"#
The foci are at #(h +-sqrt(a^2+b^2),k)#
The asymptotes are: #y = +-b/a(x-h)+k#

Matching the given equation,

#x^2/64-y^2=1" [2]", #

with equation [1], we observe that #a = 8, b = 1, and h = k = 0#.

With these values the foci are at:

#(0-sqrt(8^2+1^2),0)# and #(0+sqrt(8^2+1^2),0)#

Simplify:

#(-sqrt65,0)# and #(sqrt65,0)#

The asymptotes are:

#y = -1/8(x-0)+0# and #y = 1/8(x-0)+0#

Simplify:

#y = -1/8x# and #y = 1/8x#

The graph is:

graph{(x-0)^2/8^2-(y-0)^2/1^2=1 [-10, 10, -5, 5]}