Asked by Anonymous
Analyze: foci, vertices, asymptotes, center, transverse (major) axis. Then graph.
2x^2-y^2+4x+4y-4=0
2x^2-y^2+4x+4y-4=0
Answers
Answered by
Damon
2 x^2 + 4 x = y^2 - 4 y +4
2 x^2 + 4 x = (y-2)^2
x^2 + 2 x - (1/2) (y-2)^2 = 0
x^2 + 2 x + 1 - (1/2) (y-2)^2 = 1
(x+1)^2 - (y-2)^2/(sqrt 2)^2 = 1
Hyperbola, you look up the stuff for it.
2 x^2 + 4 x = (y-2)^2
x^2 + 2 x - (1/2) (y-2)^2 = 0
x^2 + 2 x + 1 - (1/2) (y-2)^2 = 1
(x+1)^2 - (y-2)^2/(sqrt 2)^2 = 1
Hyperbola, you look up the stuff for it.
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