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Write an equation of the hyperbola with vertices at (6,-3)(6,1) and foci at (6,-6)(6,4)
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@ganeshie8 so i solved this o ut and for my final answer i got (x-6)^2/16-(y+2)^2/12=1
the center is (6,-1) a=4, c=2? opens up and down
it is a vertical hyperbola right ?
so y^2 term must come first
ok
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|dw:1397824503908:dw|
\(\large \dfrac{(y-(-1))^2}{a^2} - \dfrac{(x-h)^2}{b^2} = 1\)
\(\large a\) = distance between center and vertex = 1 - -1 = 2
right ?
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