How do you find the vertices, asymptote, foci and graph #100x^2-81y^2=8100#?

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Answer 1 · 2016-11-01T19:52:07

Please see the explanation.

The standard form for this kind of hyperbola is:

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

where:

#(h, k)# is the center

#(h-a, k)# and #(h+a, k)# are the vertices

#(h - sqrt(a^2 + b^2), k)# and #(h + sqrt(a^2 + b^2), k)# are the foci

and the asymptotes are

#y = b/a(x - h) + k# and #y = -b/a(x - h) + k#

Given:

#100x^2 - 81y^2 = 8100#

Divide both sides by 8100:

#x^2/81 - y^2/100 = 1 #

Write the denominators as squares and add -0 inside the numerator squares:

#(x-0)^2/9^2 - (y-0)^2/10^2 = 1 #

Now, the vertices can be written by observation #(-9, 0)# and #(9 , 0)#

The foci are written with at trivial computation of #sqrt(10^2 + 9^2) = sqrt(181)#:

#(-sqrt(181), 0) and (sqrt(181), 0)#

The asymptotes are:

#y = 10/9x# and #y = -10/9x#