What is the intersection of $xy+yz+xz $ and $x^2+y^2-z^2$? Where $x,y,z \in \Bbb R$. It looks very difficult to found a parametrization :S
parametrization of the intersection of two surfaces
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2@Daniel: Then please clarify the question. People shouldn't have to read through the comments in order to understand the question. There's an edit link underneath the question. – 2012-11-09
2 Answers
With $x=r\cos\phi$ and $y=r\sin\phi$, the equations become $r^2\cos\phi\sin\phi+rz(\cos\phi+\sin\phi)=0$ and $r^2-z^2=0$. Thus we have $r=\pm z$ and therefore $\cos\phi\sin\phi\pm(\cos\phi+\sin\phi)=0$. With $\phi=\theta+\frac\pi4$ and thus $\cos\phi=(\cos\theta-\sin\theta)/\sqrt2$ and $\sin\phi=(\cos\theta+\sin\theta)/\sqrt2$, this becomes $(\cos^2\theta-\sin^2\theta)/2\pm\cos\theta/\sqrt2=0$. Using $\sin^2\theta=1-\cos^2\theta$, solving the resulting quadratic equation for $\cos\theta$ and keeping only the solutions with $|\cos\theta|\le1$ leads to $\cos\theta=\pm(1-1/\sqrt2)$ and thus $\phi=\frac\pi4\pm\arccos(\pm(1-1/\sqrt2))$, with the inner sign determined by the sign of $z$. From this $\cos\phi$ and $\sin\phi$ can be obtained as algebraic constants, and the intersection takes the form of two lines through the origin, parametrized by $z$.
Here is another idea, although I will leave out details, and hope it only complements joriki's explicit answer.
These equations define cones in $\mathbb R^3,$ and their intersection must also form a cone. This means that given a point on the intersection, we know that the line through that point and the origin is entirely contained in the intersection. We can thus greatly simplify the question by restricting to the case $z=1.$ (We can easily see that the only point of intersection with $z=0$ is the origin.) In this case we simply need to find the intersection points (in the $xy$-plane) of the circle $x^2+y^2=1$ and the curve $x+y+xy=0.$ Once we have such intersection points, say one of them is $x=a,y=b,$ then we know that $\{(az,bz,z)\vert z\in\mathbb R\}$ is a line in the intersection. In fact, one can see that the intersection will be parametrized exactly by such lines.
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1If I am not mistaken, there are only two solutions for $x$ for the system of inhomogeneous equations ($x^2$ is solution of $t^3+t^2-1=0$). So the system is completely solved. – 2012-11-09