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

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1since 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I know the origin is at (0,0) my equation should look like: y^2/a^2x^2/b^2=1

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1the center appears to already be at 0; otherwise youd have different 'x' parts for the foci

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0jk it would be x^2/a^2y^2/b^2=1

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1the asymptote tells us what a and b are really; since it is in the form y = (b/a)x

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.12 b  x =  x 1 a the y component is b and the x compnent is a

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Right and I know I need to find c^2=a^2+b^2 I know C is 3^2 right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1x^2 y^2    = 1 1^2 2^2

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1a is only needed if you are looking to find a or b; we know both a and b from the asympotote

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok but that can't be right....I think we may have a wrong....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0my 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0x^2 y^2    = 1 9 4 I would guess this one but I'm not sure

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1we need to determine the "rectangle" of the hyperbola; to so that, lets write it as an ellipst

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1x^2 y^2  +  = 1 the vertices of this ellipse are the top and sides a^2 b^2 of your asympotoe information

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.12x 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?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1and it should end up looking like this inthe end

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok.... ugh this stupid math class their answer key never matches up to what I get lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1our y compnent is 2^2 regardless right :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.1id go with your first option; dbl chk by making y=0 and seeing what you get for an x value.. maybe :)
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