How to find the number of intersection points between affine curves with multiplicity 2?

Question

How many intersection points with multiplicity 2 are there in the intersection of the affine curves $y^{3}-2 y x+1=0$ and $3 x+2 y+3=0$ (defined over $\mathbb{C}) ?$

I cant even find an intersection point for these two curves, wolfram gives me this:

$x=\frac{1}{27}\left(-19-4 \sqrt[3]{19}+2 \times 19^{2 / 3}\right) \wedge y=\frac{1}{18}\left(19+4 \sqrt[3]{19}-2-19^{2 / 3}\right)-\frac{3}{2}$

My usual methodology is to find at least one intersection point, then find another point on the line to create a parameterization in t, and then the intersection multiplicity for the point used is the multiplicity of t in the composition of the curve and the newly parameterised line in t. I would do this for all intersection points to find how many have multiplicity 2.

Since I can only estimate an intersection point, I assume there is an alternate method using properties of affine curves that I'm not aware of. Any help on this would be greatly appreciated!

Answer 1

Differentiating, you have $3y^2y'-2(y+y'x)=0$ and $3+2y'=0$. If the multiplicity of an intersection point is $2$ or greater, then $y'$ will agree on the two curves at that point. So $y'=-3/2$, and

$$-\frac92y^2-2y+3x=0$$

From the second equation, $3x=-2y-3$. Combine this with the above equation and there are only two possibilities for $y$. Then the line equation can once again be used to pair each of those $y$-values with an $x$-value.

Then the question will be if these two ordered pairs actually lie on the first curve. If they do, further investigate their multiplicity (is is 2, or is it greater?) If they do not, then there are no such high multiplicity intersection points.

Answer 2

Alternative to my other post. You can use the second equation to solve for $x$ and reduce the first equation to $$y^3-2y\left(-\frac{2}{3}y-1\right)+1=0$$

Or:

$$3y^3 + 4y^2+6y +3=0$$

There is only one real solution to this equation, since $\frac{d}{dy}\left(3y^3 + 4y^2+6y +3\right)=9y^2+8y+6=\left(3y+\frac{4}{3}\right)^2+\frac{38}{9}>0$ implies the function is strictly increasing.

So the other two roots are a complex conjugate pair. So the three $y$-values are distinct. So the three intersection points are distinct and must each have degree $1$.