two equivalents equations with different representations in the plane

Question

Consider the parametric curve $$C:\begin{cases} x = 4e^{t/4} \\ y = 3e^{t} \\ \end{cases} $$ A cartesian equation for this curve is $y=\frac{3x^4}{256}$. The problem is that $x=4e^{t/4}$ is positive so in the plane the representation of the curve will be in the first quadrant, while the equation $y=\frac{3x^4}{256}$ is valid for all $x\in \mathbb R$. so the two curves coincide only on the first quadrant but the graph of $y=\frac{3x^4}{256}$ has another part which in the second quadrant which is symmetric to the first part by the $y-axis$. Is is possible that the two equations have different graphs in the plane ? how can we explain that ? thank you for your help !

Answer 1

There is not much to explain. You are given a map $${\bf f}:\quad {\mathbb R}\to{\mathbb R}^2, \qquad t\mapsto\bigl(4e^{t/4}, 3e^t\bigr)\ ,$$ and you have proven that all points of the set $C:={\bf f}({\mathbb R})$ satisfy the equation $3x^4-256y=0$.

Note that even the map $${\bf g}:\quad {\mathbb R}\to{\mathbb R}^2, \qquad t\mapsto\bigl(4, 3\bigr)\ ,$$ would have the property that all points of ${\bf g}({\mathbb R})$ would satisfy this equation.

It is another thing if you start with the equation $3x^4-256y=0$ and are asking for a map ${\bf h}$ that produces the solution set of this equation in a bijective way. The given ${\bf f}$ will certainly not qualify.