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The Quasilinear equation of parabola y2 = 4x is () A. x=1B. y=1C. x=-1D. y=-1
∵ parabola y2 = 4x, P2 = 44 = 1, its quasilinear equation is x = - 1
Quasilinear equation of parabola y = x ^ 2-4x + 5
Parabola y = x 2 – 4x + 5 = (x - 2) 2 + 1, that is, (x - 2) 2 = y – 1, let X '= x – 2 and y' = y – 1, we can get (x ') 2 = y', the Quasilinear equation is y '= - 1 / 4, so the original quasilinear equation of parabola is y – 1 = - 1 / 4, that is, y = 3 / 4
The Quasilinear equation of parabola y = 4x2 is______ ．
By sorting out the parabolic equation, we get that x2 = 14y, ∵ P = 18 ∵ the opening of parabolic equation is upward, ∵ quasilinear equation is y = - 116, so the answer is: y = − 116
The Quasilinear equation of parabola y = 4x2 is______ ．
By sorting out the parabolic equation, we get that x2 = 14y, ∵ P = 18 ∵ the opening of parabolic equation is upward, ∵ quasilinear equation is y = - 116, so the answer is: y = − 116
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