How do you find the equation for a hyperbola centered at the origin with a vertical transverse axis of lengths 12 units and a conjugate axis of length 4 units?

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Answer 1 · 2017-12-15T21:22:28

The general form for a the equation of a hyperbola with a vertical transverse axis is:

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

The center of the general form is #(h,k)#. We are told that the center is the origin: #(0,0)#, therefore, we can remove h and k from the equation:

#y^2/a^2-x^2/b^2=1#

The length of the transverse axis is #2a#

#2a = 12#

#a = 6#

#y^2/6^2-x^2/b^2=1#

The length of the conjugate axis is #2b#:

#2b = 4#

#b = 2#

#y^2/6^2-x^2/2^2=1#

Done.