How do you write the hyperbola #36x^2-9y^2=324# in standard form?

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Answer 1 · 2016-10-09T13:16:37

#(6x)^2/18^2-(3y)^2/18^2=1#

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

Notice that #36=6^2"; "9=3^2"; "324=18^2#

So we can write:

#(6x)^2/a^2-(3y)^2/b^2=18^2#

So #h" and "k=0 larr # symmetrical about the origin

Divide both sides by #18^2#

#(6x)^2/18^2-(3y)^2/18^2=1#

Tony B