Equations of Projection of a curve to a plane This is inspired by an exercise in Hartshorne, but really I don't understand how to do this in general.
Suppose I have some system of equations that defines a curve, and some point, either on the curve or not, and I want to project the curve from this point to some hyperplane. How can I calculate this? I suppose if we can also project surfaces to hyperplanes or just varieties in general, we can do the same kind of thing to the family of equations, no?
If the curve is given parametrically, I can do this with just some easy algebra, but what if the curve is given implicitly by some equations? Here is the example exercise in Hartshorne:
Ex I.5.11: $\textit{The Elliptic Quartic Curve in } \mathbb{P}^3:$ Let $Y$ be the algebraic set defined by the equations $x^2 - xz - yw = 0$ and $yz - xw - zw = 0$. Let P be the point $[x:y:z:w] = [0:0:0:1]$, and $\phi$ be the projection from $P$ to the plane $w = 0$. Show that $\phi$ induces an isomorphism of $Y-P$ with the plane cubic $y^2z - x^3 + xz^2$.
I am reasonably sure if I can see how to compute what the projection does to the equation of the curve, I can see the isomorphism. But maybe this is not the way to go and even then I won't be able to complete the exercise.
I suppose my question is twofold: first, how do you carry out these kinds of calculations. Second, if doing such a calculation will not help me see the isomorphism, what is the best way to go about this exercise?
 A: I have done this in two cases now. One of them was another Hartshorne exercise which I must have skipped, otherwise I'd have surely ran into this problem back then. This will serve as a nice toy case.
Exercise 4.4c of Hartshorne asks one to show there is a birational map from the curve $y^2z = x^2(x+z)$ to $\mathbb{P}^1$ given by projection as above, but with $3$ variables instead of $4$. The projection map obviously sends $[x:y:z] \mapsto [x:y]$, which is valid when at least one of $x$ and $y$ is non-zero, an open set, but it is not clear what the inverse map should be. To figure it out, we exploit the equation.
$y^2z = x^3 + x^2 z \implies (y^2 - x^2)z = x^3 \implies z = \frac{x^3}{y^2 - x^2}$. This is not going to be defined along the lines $y = \pm x$, but since we're only dealing with rational maps, it will be defined on the open set which is the complement of these two lines. So on the complement of these lines, we can write the inverse as $[x:y] \mapsto [x:y:z] = \big[x:y:\frac{x^3}{y^2 - x^2}\big ] = [(y^2 - x^2)x : (y^2 - x^2)y, x^3]$. 
This is a good example because the proof of Corollary I.4.5 informs us that such birational maps can be turned into isomorphisms of open sets, which is what we'll want in the bigger case, which we'll now go through.
The curve is defined by the equations $x^2 - xz - yw = 0$ and $yz - xw - zw = 0$. The projection obviously sends $[x:y:z:w] \mapsto [x:y:z]$. To figure out the image of the curve (and this is the strategy generically), we eliminate $w$.
The first equation implies $w = \frac{x^2 - xz}{y}$ while the second implies $w = \frac{yz}{x+z}$. Equating these eliminates $w$ and gives the desired curve $y^2z - x^3 + xz^2 = 0$. This equality holds when $y \neq 0$. Staring at the equations for a moment informs us that $y = 0$ gives the missing point $[1:0:-1]$ which corresponds with $[0:0:0:1]$. So this gives a pair of maps between the desired sets. These can be shown to be morphisms by direct calculation - one follows from Exercise I.3.14, and the other follows from staring at the equations. We end up with equations $[x:y:z] \mapsto [xy:y^2:yz:x^2 - xz]$ and $[x,y,z] \mapsto [(x+z)x:(x+z)y:(x+z)z:yz]$, which are both easily seen to suffice.
