Find the standard form of the equation of the hyperbola with the given characteristics. foci: (+-3,0) Asymptotes: y=+-2x

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Find the standard form of the equation of the hyperbola with the given characteristics. foci: (+-3,0) Asymptotes: y=+-2x

Mathematics
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since the foci are located parallel to the x axis; we can make a note of that and say that they open to the left and right
I know the origin is at (0,0) my equation should look like: y^2/a^2-x^2/b^2=1
the center appears to already be at 0; otherwise youd have different 'x' parts for the foci

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jk it would be x^2/a^2-y^2/b^2=1
the asymptote tells us what a and b are really; since it is in the form y = (b/a)x
2 b -- x = --- x 1 a the y component is b and the x compnent is a
Right and I know I need to find c^2=a^2+b^2 I know C is 3^2 right?
x^2 y^2 --- - --- = 1 1^2 2^2
a is only needed if you are looking to find a or b; we know both a and b from the asympotote
ok but that can't be right....I think we may have a wrong....
my options are x^2 y^2 --- - --- = 1 9 4 x^2 y^2 --- - --- = 1 9/5 36/5 x^2 y^2 --- - --- = 1 36/5 9/5
x^2 y^2 --- - --- = 1 9 4 I would guess this one but I'm not sure
lets recheck then :)
we need to determine the "rectangle" of the hyperbola; to so that, lets write it as an ellipst
uh.. ok?
x^2 y^2 --- + ---- = 1 the vertices of this ellipse are the top and sides a^2 b^2 of your asympotoe information
correct
2x is the asympy.... this tells us the from the center, we go up 2 and over 1 to for the top right corner of the rectangle; and our ellipse will be 2 high and 1 left right?
right
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ok I got it
and it should end up looking like this inthe end
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ok.... ugh this stupid math class their answer key never matches up to what I get lol
our y compnent is 2^2 regardless right :)
id go with your first option; dbl chk by making y=0 and seeing what you get for an x value.. maybe :)
ok thanks

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