Parabolas intersection

Question

We can get parabolic equation by using only focus and diretrix of parabola: $y^2 = 2px$, where p is a shortest distance between focus and directrix. But this equation defines parabola in coordinate system where directrix is parallel to OY axis, focus have coordinates $(p/2, 0)$ and parabola's vertex have coordinates $(-p/2, 0)$. So we have parabola in a special coordinate system and if we want to intersect this parabola with line we need to rotate and shift line to got it in parabola's coordinate system then find intersection and finally shift and rotate intersections back to our base coordinate system.

But what if we need to find intersections of two parabolas? They can intersects in up to 4 points, but it seems to be hard to solve quartic equation... Where can I get information about this case?

Thanks for help.

Welcome to math.stackexchange.com! I have edited your post a bit to improve its look. E.g., y^2 = 2px becomes the displayed formula if you enclose it in $ signs (TeX). — Hans Engler, Apr 16 '14 at 12:43
You can get point of intersection without rotating axes. Simple substitution. — evil999man, Apr 16 '14 at 13:04
And you can't run from that quartic. If you want to find intersection of 2 parabolas, you gotta solve it(even making it is not a piece of cake) — evil999man, Apr 16 '14 at 13:06
@Awesome can you explain how I can get intersection without rotating? Without it, two parabolas with equations like $y^2 = 2px$ even can't have 4 points of intersection. Or I misunderstand you? — plomovtsev, Apr 16 '14 at 13:45
@mopdobopot I was talking about intersection of line with parabola — evil999man, Apr 16 '14 at 13:47
@Awesome sorry, it's not clear for me how it can be possible, could you explain it? :) — plomovtsev, Apr 16 '14 at 13:53
$$L : y=mx+n$$$$P : ax^2+by^2+2gx+2fy+2hxy+c=0$$ Substitute y from L in P. Solve the quadratic — evil999man, Apr 16 '14 at 13:55
@Awesome thank you, now I understood! But I find rotation-way easier than calculating terrible coefficients for original parabolic equation :) — plomovtsev, Apr 16 '14 at 14:36
It ain't that hard if good values are given. Else, it WILL get dirty. — evil999man, Apr 16 '14 at 15:32
@Awesome said “And you can't run from that quartic.” But I would think otherwise: my preferred method for intersecting conics will handle this case as well, and it only needs to solve quadratic and cubic equations. Of course, one could argue that in effect I'm still solving a quartic equation, only in a different way. — MvG, Apr 17 '14 at 12:22

score 1 · Accepted Answer · edited Apr 13 '17 at 12:20

Suppose you have the focus at point $(a, b)$ and a directrix with equation $cx+dy=e$. Then a point $(x,y)$ is on the parabola if it has the same distance from focus and directrix, which means

$$(a-x)^2+(b-y)^2=\frac{(cx+dy-e)^2}{c^2+d^2}$$

You can rewrite that as

$$ (x,y,1) \begin{pmatrix} d^2 & -cd & ce-a(c^2+d^2) \\ -cd & c^2 & de-b(c^2+d^2) \\ ce-a(c^2+d^2) & de-b(c^2+d^2) & (a^2+b^2)(c^2+d^2) - e^2 \end{pmatrix} \begin{pmatrix} x \\ y \\ 1 \end{pmatrix} = 0 $$

If you choose the Hesse normal form for the line, with unit length normal vector, you have $c^2+d^2=1$ which will make the matrix easier to read. Note that the entries of this matrix correspond to the coefficients Awesome assumed in his comment.

Do this for both your parabolas, and you get two matrices which you can then feed into the generic method for intersecting conics.

Parabolas intersection

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