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What are the vertices of the hyperbola given by the equation (y4)^2/121  (x+9)^2/141 = 1?
 one year ago
 one year ago
What are the vertices of the hyperbola given by the equation (y4)^2/121  (x+9)^2/141 = 1?
 one year ago
 one year ago

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satellite73Best ResponseYou've already chosen the best response.1
first you need the center so you know that one ?
 one year ago

dricboylanBest ResponseYou've already chosen the best response.0
No, I really don't understand this section which is why I was asking for help.
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
center of \[\frac{(yk)^2}{a^2}\frac{(xh)^2}{b^2}=1\]is \((h,k)\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
in your case it is \((9,4)\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
because the \(y\) part comes first (before the minus sign) this hyperbola looks like this (roughly) dw:1353293873616:dw
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
to find the vertices, go \(a\) units up up and down from the center. that is why you needed the center first in your case \(a^2=121\) and so \(a=11\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
since the center is \((9,4)\) if you go up 11 units you are at \((9,15)\) and if you go down 11 units you are at \((9,7)\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
hope that is clear, and also that you see it is not that hard once you know what you are doing
 one year ago

dricboylanBest ResponseYou've already chosen the best response.0
I'm still extremely lost, as you can tell math isn't my strong point. I appreciate the help however.
 one year ago
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