Can a quartic polynomial curve in the two dimensional affine space over complex numbers have 4 parallel asymptotes?
I'm studying plane curves in affine and projective spaces over real and complex numbers. The question is: can a quartic polynomial curve in the affine space over the complex numbers have 4 different asymptotes, all parallel to the same line, $r : x  2y + 3 = 0$?
To my understanding, asymptotes in an affine curve are tangents of improper points of the curve, that is, if we immerge the affine space into the projective one, we lose a plane that contains all of the directions of the lines in the affine space itself. Let's call the projective coordinates $x_0, x_1, x_2, ...$,then we choose an immersion that gives us the improper plane $\{x_0 = 0\}$. So the improper point of the curve can be found by homogenizing the equation of the line itself , and finding the improper points, which are the same for all 4 lines since they're all parallel to $r$, $P = [0, 1, 2]$. So, $P$ needs to have 4 different tangents, and that is not a contradiction per se. Since, by dehomogenizing the line equations, we need to get something like $x  2y + c, c$ a complex number, I guess we'd need to factor the tangent cone in $P$ as something like
$(x_1  2x_2 + c_1x_0)(x_1  2x_2 + c_2x_0)(x_1  2x_2 + c_3x_0)(x_1  2x_2 + c_4x_0)=0$.
How should I be able to do that? Also, did I miss something, am I able to do that in the first place?
Thank you so much in advance.
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Questions at this level should come with a good bounty.
I really can’t afford to give money away right now since I barely have money for myself. It’s fine if nobody answers, I don’t expect strangers to take time off to help me, know it’s unlikely I’ll get an answer, but my exam is a few days away, I tried to ask this same question elsewhere to no avail, and I don’t know anyone who could know something like this, so I though I might as well give this a try.