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Mathematics 19 Online
OpenStudy (anonymous):

Find the vertices and asymptotes of the hyperbola. 25y^2 - 16x^2 = 400

OpenStudy (anonymous):

\[y^2/(400/25) - x^2/(400/16) =1\]

OpenStudy (anonymous):

\[y^2/(4)^2- x^2/(5)^2= 1\]

OpenStudy (anonymous):

shouldnt it be 25/400 and not the other way around

OpenStudy (anonymous):

no 25 is multiplied with y^2

OpenStudy (anonymous):

y=(+/-)b/a x are the asymptotes

OpenStudy (anonymous):

ok so the asymptote.. y=+-4/5x

OpenStudy (anonymous):

then does that mean the vertices are (0,+-4)

OpenStudy (anonymous):

Yes uxma but as you are taking 400 from the left hand side it should be the denominator in this case

OpenStudy (anonymous):

oh..i made a mistake, the axis of hyperbola is y axis

OpenStudy (anonymous):

right ishan, n then 25 comes in the denominator of 400 :)

OpenStudy (anonymous):

So am i still right then? y=+-4/5x and (0,+-4)

OpenStudy (anonymous):

So am i still right then? y=+-4/5x and (0,+-4)

OpenStudy (anonymous):

right :)

OpenStudy (anonymous):

sweet thanks

OpenStudy (anonymous):

54 to99 simplest form

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