SOLUTION: what is the conic, center, and intercepts of 8x^2-4y^2=16?

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Question 197717: what is the conic, center, and intercepts of 8x^2-4y^2=16?
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
what is the conic, center, and intercepts of 8x^2-4y^2=16?
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Put in standard form:
Divide thru by 16 to get:
8x^2/16 - 4y^2/16 = 1
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Rearrange:
x^2/2 - y^2/4 = 1
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conic: hyperbola
center: (0,0)
intercepts:
Let y = 0 to find x-intercepts:
x^2/2 = 1
x^2 = 2
x = +sqrt(2), x = -sqrt(2)
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There are no y-intercepts.
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Cheers,
Stan H.