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

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

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

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

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

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1because the \(y\) part comes first (before the minus sign) this hyperbola looks like this (roughly) dw:1353293873616:dw

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1to 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 \(a=11\)

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1since 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 \((9,7)\)

satellite73
 3 years ago
Best ResponseYou've already chosen the best response.1hope that is clear, and also that you see it is not that hard once you know what you are doing

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