There was no reasoning But there was a warning that I like mathsOriginally Posted bysthomas

Let P0 be 2-dimensional point [P0.x; P0.y]

Consider Bezier curve with control points P0, P1, P2 and P3:

x(t) = P0.x*(1-t)^3 + 3*P1.x*t(1-t)^2 + 3*P2.x*t^2(1-t) + P3.x*t^3

y(t) = P0.y*(1-t)^3 + 3*P1.y*t(1-t)^2 + 3*P2.y*t^2(1-t) + P3.y*t^3

where 0 <= t <= 1

Add the constraint imposed by the tangential definition: the points P0, P1, P2 and P3 are equally spaced in x direction:

P0.x = a P1.x = a + b P2.x = a + 2b P3.x = a + 3b

Put into equation for x(t):

x(t) = a*(1-t)^3 + 3*(a + b)*t(1-t)^2 + 3*(a + 2b)*t^2(1-t) + (a + 3b)*t^3

= a { (1-t)^3 + 3*t(1-t)^2 + 3*t^2(1-t) + t^3 } + b { 3t(1-t)^2 + 6*t^2(1-t) + 3*t^3 }

= a { ((1-t)+t)^3 } + b { 3t (1 - 2t + t^2) + 6(t^2 - t^3) + 3*t^3 }

= a { 1^3 } + b { 3t - 6t^2 + 3t^3 + 6t^2 - 6t^3 + 3*t^3 }

= a + b { 3t }

= 3t*b + a

Therefore,

x = 3t*b + a

x - a = 3t*b

t = (x - a)/(3b)

Now we can substitute 't' into y(t) = P0.y*(1-t)^3 + 3*P1.y*t(1-t)^2 + 3*P2.y*t^2(1-t) + P3.y*t^3

That would be a lot of algebra, but since t is linearly dependent on x will can see that y will be cubic in terms of x.

Hence the tangential form can only express curves in form y = f(x).