How do you write a standard form equation for the hyperbola with vertices at (-2, 0) and (2, 0) and foci at (-6, 0) and (6, 0)?

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Answer 1 · 2017-05-26T20:29:39

Because the vertices are horizontal, we know that the standard form is,

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

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

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

Using the form of the vertices and the given vertices we can write the following equations:

#-2 = h-a#
#2 = h+a#
#k = 0#

Solving the first two equations we have:

#h = 0#
#a = 2#
#k =0#

Using the form of the foci and one of the given foci we can write:

#6 = h + sqrt(a^2+b^2)#

Substitute in the known values for h and a:

#6 = 0 + sqrt(2^2+b^2)#

Solve for b:

#36=4+b^2#

#b = sqrt32#

Substituting the know values into equation [1]:

#(x-0)^2/2^2-(y-0)^2/(sqrt32)^2=1" [2]"#

Equation [2] is the answer.