How do you find the equation of the hyperbola with center at the origin and vertices #(+-1,0)# and asymptotes #y=+-3x#?

1 Answer
Douglas K.
Jan 2, 2017

Please see the explanation.

Explanation:

Please observe that the vertices are horizontally oriented, #(-1, 0) and (1,0)#, therefore, the hyperbola is the horizontal transverse axis type . From the reference, the standard Cartesian form for the equation of a hyperbola with a horizontal transverse axis is:

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

where h and k are the center point #(h,k)#, "a" is the distance from the center to a vertex, and #+-b/a# is the slope of asymptotes.

We are given that the center is the origin so substitute 0 for h and k into equation [1]:

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

We can see that the distance from the center to either vertex is 1, therefore substitute 1 for a into equation [2]:

#(x - 0)^2/1^2 - (y - 0)^2/b^2 = 1" [3]"#

We are given that the slope of the vertices is #+-3#; this allows us to solve for b:

#+-b/a = +-3#

Substitute 1 for a:

#+-b/1 = +-3#

#b = 3#

To substitute 3 for b into equation [3]:

#(x - 0)^2/1^2 - (y - 0)^2/3^2 = 1" [4]"# Answer

You can simplify equation [4] if you like, but I recommend that you leave it in this form.

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