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

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satellite73 Group TitleBest ResponseYou've already chosen the best response.1
first you need the center so you know that one ?
 2 years ago

dricboylan Group TitleBest ResponseYou've already chosen the best response.0
No, I really don't understand this section which is why I was asking for help.
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
center of \[\frac{(yk)^2}{a^2}\frac{(xh)^2}{b^2}=1\]is \((h,k)\)
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
in your case it is \((9,4)\)
 2 years ago

satellite73 Group TitleBest 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
 2 years ago

satellite73 Group TitleBest 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 \(a=11\)
 2 years ago

satellite73 Group TitleBest 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 \((9,7)\)
 2 years ago

satellite73 Group TitleBest 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
 2 years ago

dricboylan Group TitleBest 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.
 2 years ago
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