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Mathematics 20 Online
OpenStudy (anonymous):

What is the equation, in standard form, of a vertical hyperbola with asymptotes at y + 8 = ±2/9(x – 8)?

OpenStudy (amistre64):

square the slope

OpenStudy (amistre64):

this is your a^2 b^2 parts

OpenStudy (amistre64):

pull the center from the equation given of the asymptotes

OpenStudy (amistre64):

since vertical is up and down, y is +, and x is -

OpenStudy (amistre64):

pfft, just use the x and y parts inthe hyper :)

OpenStudy (amistre64):

y + 8 = ±2/9(x – 8) 4/81 and since y/x defines slope, then 4 is under ys and 81is under xs (y+8)^2 (x-8)^2 ------- - -------- = 1 4 81

OpenStudy (anonymous):

Thank you!

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