Projections of the standard twisted cubic on $\mathbb{P}^2$ (Harris 3.8) As in the title, I am trying to find the equation of the projection of the twisted cubic $$\mathscr{C}=V(Z_0Z_2=Z_1^2, \quad Z_0Z_3=Z_1Z_2,\quad Z_1Z_3=Z_2^2)\subset\mathbb{P}^3$$ 
on an hyperplane $\mathbb{P}^2$ from the points $p=[1:0:0:1]$ and $p=[0:0:1:0]$. 
Since worjing with the resultant of polynomials is computationally inefficient my idea is the following, let us consider the second case, which I think is neater:  


*

*Call $X=\pi(\mathscr{C})$ the projection. Fix $p=[0:0:1:0]$ and $\mathbb{P}^2=V(Z_2)$, with coordinates $Z_0,Z_1,Z_3$. 

*Then $q=(z_0,z_1,z_3)\in X$ if and only if the line $\bar{pq}$ meets $\mathscr{C}$.
I translate this condition as:
$$\exists\lambda\neq 0\ | (z_0,z_1,\lambda,z_3)\in\mathscr{C}$$

*Then I plug this condition in the equations and develope the computation trying to delete $\lambda$ to obtain the desired equation 


Then I ask:


*

*Is the approach correct? Is the condition I fixed correct?

*I believe the other case is analogous and I would do it it by chosing $\mathbb{P}^2=V(z_0+z_3)$, is it the right idea?

*The excercise then asks to show that any other projection is projectvely equivalent to one of the two above: I have no idea about how to prove this. 

 A: Yes, your approach is correct (it may be easier to go the other way around, though - pick a point $c=[s^3:s^2t:st^2:t^3]\in C$ and compute when $pc$ intersects $V(Z_2)$). The condition you've fixed is correct for the first case.
For the second case, you should instead fix $\Bbb P^2=V(Z_1-Z_3)$: there's most of a worked example (with the same sign error from the asker, even!) here.
For the third case, the answer depends on doing the computations from part (1) first. Since your post doesn't indicate that you've finished these yet, I'm going to spoiler-text the remainder of the answer so you don't accidentally ruin the fun for yourself:

 As you've calculated from 1), the equations of the cubic curves you get from these projections are $X_1^3=X_0^2X_2$ and $X_1^3=X_2^3+X_0X_1X_2$. What do you notice about these curves? Are they smooth?

With that hint, we proceed:

 Both curves are singular (the first is the cuspidal cubic, the second is the nodal cubic). This suggests that we use the fact that there are only two singular irreducible projective plane cubic curves up to projective equivalence. So we need to prove that every projection of the twisted cubic is singular. The strategy for this is to show that if we project from a point on a tangent line to the twisted cubic, we get a cuspidal cubic, whereas if we project from a point on a secant line, we get the nodal cubic. This covers all the cases, since every point in $\Bbb P^3$ is on either a secant or a tangent line of the twisted cubic. I'm going to leave the computations to you so as to not spoil the fun :)

I encourage you to follow through and do the calculations yourself - this can be a really good learning experience!
