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 de-homogenizing 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.
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 de-homogenizing 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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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.