# Curves intersecting at the origin

1. Apr 6, 2005

### b0mb0nika

i have to show how many times the curves intersect at the origin

y^4 = x^ 3 and x^2y^3 - y^2+ 2x^7= 0

i dont really know how to start solving this :uhh:

2. Apr 6, 2005

### b0mb0nika

i think the answer is 6
could anyone tell me if it's right?

3. Apr 7, 2005

### shmoe

It looks right. You found 22 solutions (over the complex numbers) with x and y non-zero? No solutions for the points at infinity in projective space?

4. Apr 7, 2005

### matt grime

I plotted them in surf. at the origin one is a U but turned on it's side with the "opening" to the right. the other is a cusp, or locally < shaped and the point of the < meets the rounded bottom of the U, I don't know how many "crossings" you'd wish to count that as.

5. Apr 9, 2005

### b0mb0nika

shmoe.. i didnt actually find the solutions.
i let t^3=y from the frist equation
and then expressed everything in terms of t in the second equation, and then by
a theorem ( which i dont know the name of).. the lowest power of the non-zero
terms is the number of time the curves intersect at 0.

6. Apr 9, 2005

### shmoe

Hmm, I'm not sure of the theorem you're using, could you give it's statement or a reference to where it can be found? I'd like to have a look. My idea was to use Bezout's theorem to count the total number of intersections (28). You can replace the 2x^7 with 2x^4y^4, making the second equation a quadratic in the variable x^2. You can then find all non-zero solutions in complex projective space (22 in total) and show the curves intersect only once at each of these, so the origin would have 6 intersections. I admit to being pretty ignorant on most things algebraic though, so I could be way off.

7. Apr 10, 2005

### b0mb0nika

ok this is how the theorem goes:

let y = p(x) and g(x,y) = 0 be 2 curves. Assume y = p(x) contains 0= (0,0) and that (y-p(x)) does not divide g(x). Then the intersection multiplicity at 0( i assume I_0 .. I sub zero...means that) of y- p(x) and g(x,y) is the smallest degree of any non zero term of g(x,p(x)) .

i'm not sure where this came from, I missed the day when it was thought in class so I got the notes from somebody and I found it there. I dont think its in the book that we use right now for the course cause I was looking for it.

8. May 1, 2005

### HallsofIvy

Staff Emeritus
How about writing them in polar coordinates?
y^4 = x^ 3 becomes r4sin4&theta;= r3cps&theta; or r sin4&theta;= cos3&theta;. Taking r= 0 we have cos&theta;= 0, $\theta= \frac{\pi}{2}$ and $\fra{3\pi}{2}$.
The graph crosses through the origin twice.

x^2y^3 - y^2+ 2x^7= 0 becomes r5cos2&theta;sin3&theta;- r2sin3&theta;+ r7cos7&theta;= 0 or r3(cos2&theta;+ r2cos7&theta;)- sin3&theta;= 0. Taking r= 0, sin3&theta;= 0 so
sin&theta;= 0. $\theta= 0$ or $\theta= \pi$ or $\theta= 2\pi$. The graph crosses the origin 3 times (notice that 0 and $2\pi$ are different!).

9. May 2, 2005

### shmoe

I don't think that works, the number of "crossings" is not the same as the intersection multiplicity. y^4=x^3 has a zero of multiplicity 3 at the origin, the other graph a zero of order 2 (the multiplicity of the zero is the lowest power term).

Sorry I forgot about this post. This makes sense, I've seen this before in the restricted case of p(x) being linear.