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Find the vertices and asymptotes of the hyperbola. 25y^2 - 16x^2 = 400
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\[y^2/(400/25) - x^2/(400/16) =1\]
\[y^2/(4)^2- x^2/(5)^2= 1\]
shouldnt it be 25/400 and not the other way around
no 25 is multiplied with y^2
y=(+/-)b/a x are the asymptotes
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ok so the asymptote.. y=+-4/5x
then does that mean the vertices are (0,+-4)
Yes uxma but as you are taking 400 from the left hand side it should be the denominator in this case
oh..i made a mistake, the axis of hyperbola is y axis
right ishan, n then 25 comes in the denominator of 400 :)
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So am i still right then? y=+-4/5x and (0,+-4)
So am i still right then? y=+-4/5x and (0,+-4)
right :)
sweet thanks
54 to99 simplest form
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