anonymous
  • anonymous
Find the vertices and asymptotes of the hyperbola. 25y^2 - 16x^2 = 400
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[y^2/(400/25) - x^2/(400/16) =1\]
anonymous
  • anonymous
\[y^2/(4)^2- x^2/(5)^2= 1\]
anonymous
  • anonymous
shouldnt it be 25/400 and not the other way around

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anonymous
  • anonymous
no 25 is multiplied with y^2
anonymous
  • anonymous
y=(+/-)b/a x are the asymptotes
anonymous
  • anonymous
ok so the asymptote.. y=+-4/5x
anonymous
  • anonymous
then does that mean the vertices are (0,+-4)
anonymous
  • anonymous
Yes uxma but as you are taking 400 from the left hand side it should be the denominator in this case
anonymous
  • anonymous
oh..i made a mistake, the axis of hyperbola is y axis
anonymous
  • anonymous
right ishan, n then 25 comes in the denominator of 400 :)
anonymous
  • anonymous
So am i still right then? y=+-4/5x and (0,+-4)
anonymous
  • anonymous
So am i still right then? y=+-4/5x and (0,+-4)
anonymous
  • anonymous
right :)
anonymous
  • anonymous
sweet thanks
anonymous
  • anonymous
54 to99 simplest form

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