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Question 888443: What are the asymptotes and foci of the hyperbola described by -4x^2 +2y^2 =8 ?
What I did was: -4x^2/8 + 2y^2/8 = 1
Then: -x^2/2 + y^2/4 = 1
By the looks of this equation I thought that the hyperbola was vertical.
Then what I did was : a= √2 b= 2
-> y= bx/a and y= -bx/a so -> y=2x√2/ 2 (Removed the root from the denominator) and y=-2x√2/2
Now I had to find the foci:
c^2= a^2 + b^2 -> c^2= 2+4= 6 -> c=√6
Since the hyperbola is vertical this is how the foci should look like:
(0,c) and (0, -c) -> (0, √6) and (0, -√6)
Is my solution correct?
Thank you for your time.
Answer by richwmiller(17219)   (Show Source):
You can put this solution on YOUR website! hyperbola
foci | ((0, -sqrt(6)) | (0, sqrt(6)))=((0, -2.44949) | (0, 2.44949))
vertices | (0, -2) | (0, 2)
center | (0, 0)
semimajor axis length | 2
semiminor axis length | sqrt(2)=1.41421
focal parameter | sqrt(2/3)=0.816497
eccentricity | sqrt(3/2)=1.22474
asymptotes | y = -sqrt(2) x | y = sqrt(2) x
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