anonymous
  • anonymous
find the standard form of the hyperbola with v( plus minus 1, 0) through point (2, 13)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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amistre64
  • amistre64
vertices are (-1,0), and (1,0); its a convenient thing this is along the x axis i spose
amistre64
  • amistre64
(2)^2 - (13)^2 = 1 is our goal i think
amistre64
  • amistre64
4 196 --- - --- = 1 hmmm a^2 b^2

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amistre64
  • amistre64
a^2 = 1 since 1/1 = 1 for out vertice
amistre64
  • amistre64
4 - 196/b^2 = 1 4b^2 -196 = b^2 3b^2 -196 = 0 b^2 = 196/3 x^2 - 3y^2/196 = 1 is te best i can do
amistre64
  • amistre64
lol.... 13^2 = 169 ..... i always gets those confused
amistre64
  • amistre64
\[x^2 - \frac{3y^2}{169}=1\] should be better

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