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find the standard form of the hyperbola with v( plus minus 1, 0) through point (2, 13)
 2 years ago
 2 years ago
find the standard form of the hyperbola with v( plus minus 1, 0) through point (2, 13)
 2 years ago
 2 years ago

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amistre64Best ResponseYou've already chosen the best response.0
vertices are (1,0), and (1,0); its a convenient thing this is along the x axis i spose
 2 years ago

amistre64Best ResponseYou've already chosen the best response.0
(2)^2  (13)^2 = 1 is our goal i think
 2 years ago

amistre64Best ResponseYou've already chosen the best response.0
4 196    = 1 hmmm a^2 b^2
 2 years ago

amistre64Best ResponseYou've already chosen the best response.0
a^2 = 1 since 1/1 = 1 for out vertice
 2 years ago

amistre64Best ResponseYou've already chosen the best response.0
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
 2 years ago

amistre64Best ResponseYou've already chosen the best response.0
lol.... 13^2 = 169 ..... i always gets those confused
 2 years ago

amistre64Best ResponseYou've already chosen the best response.0
\[x^2  \frac{3y^2}{169}=1\] should be better
 2 years ago
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