How do you find the vertices and foci of #x^2/121-y^2/4=1#?

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Answer 1 · 2016-11-21T02:17:26

Please see the explanation.

Standard form for a hyperbola with a horizontal transverse axis is:

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

Center: #(h, k)#
Vertices: #(h - a, k) and (h + a, k)#
Foci: #(h - sqrt(a^2 + b^2), k) and (h + sqrt(a^2 + b^2), k)#

Put the given equation in standard form:

#(x - 0)^2/11^2 - (y - 0)^2/2^2 = 1#

vertices: #(-11, 0) and (11, 0)#

#sqrt(a^2 + b^2) = sqrt(11^2 + 2^2) = sqrt(125) = 5sqrt(5)#

Foci: #(-5sqrt(5), 0) and (5sqrt(5), 0)#