Considering the Cartesian form of the cubic equation: $$y(x)= ax^3 + bx^2 + cx + d $$
And considering the parametric equation of this above equation:
$$x(t)=et^3 + ft^2 + gt + h \\ y(t)=pt^3 + qt^2 + rt + s$$
Related to these equations I have the following questions:
- Is it really correct to consider these two equations are denoting the same shape? i.e by choosing wisely the parameters, can we define the same shape with both equations?
- Is it possible to convert from one form to the other? i.e I have the 8 parameters (e, f, g, h, p, q, r, s) and is it somehow possible to convert this to the Cartesian form and find the params (a, b, c, d) such that they define the same shape? note that the higher order parameters 'a' and 'b' cannot be zero.
This is related to the question: Parametric form of a cubic function
In this question : Why this parametric function for an explicit cubic form? the answer states that the we cannot convert from one form to the other without the coefficients of square and cubic term (in our case e, f, p, q ) being null or zero. If this is correct, what is the reason behind this?
The graph of the function
$$y(x)= ax^3 + bx^2 + cx + d $$
Is the same as the graph of the parametric curve
$$x(t)=t \\ y(t)=at^3 + bt^2 + ct + d$$
However you can only convert
$$x(t)=et^3 + ft^2 + gt + h \\ y(t)=pt^3 + qt^2 + rt + s$$
if $e=f=0$. If that's the case, the corresponding function is
$$y(x)= p((x-h)/g)^3 + q((x-h)/g)^2 + r((x-h)/g) + s $$
The reason is that the inverse of $x(t)$ when $\lnot(e=f=0)$ is non-polynomial, and so is the composition of the polynomial $t\mapsto y(t)$ by the non-polynomial $x^{-1}$.