SOLUTION: In which direction does the hyperbola defined by the equation below open? Check all that apply. (x-3)^2/4^2-(y+5)^2/2^2=1

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: In which direction does the hyperbola defined by the equation below open? Check all that apply. (x-3)^2/4^2-(y+5)^2/2^2=1       Log On


   

Question 677820: In which direction does the hyperbola defined by the equation below open?
Check all that apply.
(x-3)^2/4^2-(y+5)^2/2^2=1

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
In which direction does the hyperbola defined by the equation below open?
Check all that apply.
(x-3)^2/4^2-(y+5)^2/2^2=1
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(x-3)^2/16 - (y+5)^2/4 = 1
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Comment:l
Notice that when y = -5, you can solve for "x"
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Notice that when x = 3, you cannot solve for "x"
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Notice that the center of the hyperbola is at (3,-5)
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Since the curve cannot cross the line x = 3, it must
open to the right and to the left.
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Cheers,
Stan H.