Why does substituting t=f(x) into y=f(t) solve this system?

[greg_rack]

Homework Statement

A point $P$ describes, in a xy plane, the geometric place identified by this parametric system
$$\left\{\begin{matrix}
x=2t-3\\
y=\frac{1}{2}t^2-4t+2
\end{matrix}\right.$$
determine the cartesian equation $f(x,y)=0$ of the geometric place

Relevant Equations

none

Hi guys, I managed to solve this problem just by "rewriting" the first equation of the system as $t=f(x)$ and then substituting that in the second $y=f(t)$ equation, ending(of course) up with the sought $f(x,y)$ function.

The problem here is I didn't really understand what I have done and why, nor the analytical reasons for that, and I'd appreciate it if you could give me the grounding I'm missing on this type of problem :)

 

[FactChecker]

The relationship between x and t is 1-1, so either variable can be used as a parameter for y. You changed the parameter of y from t to x. That put the line in the x,y plane and removed the t parameter. That left you with the equation, $(1/8)(x+3)^2 - 2(x+3)+2-y = 0$ as the problem requested.

 

[SammyS]

Homework Statement:: A point $P$ describes, in a xy plane, the geometric place identified by this parametric system
$$\left\{\begin{matrix}
x=2t-3\\
y=\frac{1}{2}t^2-4t+2
\end{matrix}\right.$$
determine the cartesian equation $f(x,y)=0$ of the geometric place
Relevant Equations:: none

Hi guys, I managed to solve this problem just by "rewriting" the first equation of the system as $t=f(x)$ and then substituting that in the second $y=f(t)$ equation, ending(of course) up with the sought $f(x,y)$ function.

The problem here is I didn't really understand what I have done and why, nor the analytical reasons for that, and I'd appreciate it if you could give me the grounding I'm missing on this type of problem :)

@FactChecker gave you a good explanation.

In addition to that, it should be mentioned that it's bad practice to use the same symbol, in this case, $f$, for three different functions.

You were given $x$ as a function of $t$. You could say, $x=u(t)$. Then as @FactChecker points out, you can solve that for $t$, because $u$ is a 1 to 1 function. This gives $t=g(x)$. Note that $g$ is the inverse function of function $u$.

Similarly, you were also given $y$ as a function of $t$. You could say $y=v(t)$.

One way of combining those gives $y=v(g(x))$.

Post #2 shows a possible result for $f(x,y)$.