curve of intersection

KBriggs · Hodge Conjecture Offline Joined: 10 Jun 2009 Posts: 72

This is a homework question, so please don't do it for me. All I want is for someone to explain the general way to approach this problem - ei, how to find the intersection of curves:

Show that the curve of intersection of $x^2 + 2y^2 - z^2 + 3x = 1$ and $2x^2 + 4y^2 - 2z^2 - 5y = 0$ lies on a plane.

All I need is for someone to explain how to find the curve of intersection of two surfaces in general. My textbook doesn't show an example and we haven't covered it in class.

Is it as simple as multiplying the first one by 2, subtracting the second one and obtaining:

$6x + 5y = 2$ ? This says nothing aout z, though. It would appear to me that a system of two equations in three unknowns shold have infinite sollutions, but I have never tried a non-linear system before. Help?

I am a little unsure what that result means physically, if someone could enlighten me I would appreciate it.

**Posted:** Mon Sep 21, 2009 2:45 pm

mavropnevma

YES, it is that simple! Write what you obtained as $6\cdot x + 5\cdot y + 0\cdot z - 2 =0$ . What is this? Answer: the equation of a plane! So you proved the intersection of those two curves lies in this plane. You've done it.

**Posted:** Mon Sep 21, 2009 3:32 pm

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KBriggs · Hodge Conjecture Offline Joined: 10 Jun 2009 Posts: 72

He, go figure. Thanks Smile

**Posted:** Mon Sep 21, 2009 3:54 pm

suadnovic · New Member Offline Joined: 29 Oct 2009 Posts: 13

@mavropnevma

I don't agree with You about: 6x + 5y = 2

precise this is the proection of intersection curve of two surfaces at xOy plane. And just is a case that this is plane (with z=o).
Generally speacking, eliminating z from both initial equations gives proection of intersection curve of two surfaces at xOy plane, some curve like g(x,y)=0. And Intersection curve can be than obtained like intersection of one z surface and cilyndar with generatrise g(x,y)=0 and generating lines paralell to z-axes

**Posted:** Mon Nov 02, 2009 8:25 am

mavropnevma

The points that lie on $F(x,y,z)=0$ and also $G(x,y,z)=0$ , clearly also lie on any linear combination of $F$ and $G$ . If that happens to be a plane, it means they are coplanar.
It has nothing to do with projections, has it?

**Posted:** Mon Nov 02, 2009 10:47 am

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