If there is only one common point between a straight line with slope 1 and a parabola y ^ 2 = 4x, then the equation of the straight line is

If there is only one common point between a straight line with slope 1 and a parabola y ^ 2 = 4x, then the equation of the straight line is

Let the linear equation be y = x + B
Substituting parabolic equation
y^2=4(y-b)
y^2-4y+4b=0
The discriminant is 0, so B = 1
Linear equation y = x + 1

There is only one common point between the straight line passing through the point (0, - 1) and the square of the parabola y = 4x. The equation of the straight line is obtained

1. A straight line parallel to the x-axis
Namely
y=-1
2. Tangent line
Let the line be
y+1=kx
y=kx-1
Substituting Y & # 178; = 4x
k²x²-2kx+1=4x
k²x²-(2k+4)x+1=0
1)k=0
Already asked
2)k≠0
Δ=4k²+16k+16-4k²
=16k+16=0
k=-1
The straight line is
y=-x-1
3. Straight line without slope
That is, x = 0

The Quasilinear equation of parabola y = 2x2 is______ ．

The equation of parabola can be changed to x2 = 12Y, so p = 14. The Quasilinear equation is y = - 18, so the answer is y = - 18