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OpenStudy (anonymous):
Check my answer? Question: Write the equation of the hyperbola with a center at (6,2), vertices along the major axis at (6,6) and (6,-2), and minor axis with a length of 6 .. My answer: (x-h)^2/a^2 - (y-k)^2/b^2 =1 then (x-6)^2/0^2-(y-2)^2/4^2=1 then (x-6)^2/0-(y-2)^2/16=1
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