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What is the equation, in standard form, of a vertical hyperbola with asymptotes at y + 8 = ±2/9(x – 8)?
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square the slope
this is your a^2 b^2 parts
pull the center from the equation given of the asymptotes
since vertical is up and down, y is +, and x is -
pfft, just use the x and y parts inthe hyper :)
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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
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