How to find the equation of the surface obtained by cutting the cylinder $x^2+y^2=9$ with the plane $x-y+z=4$
I know that it should be an ellipse . what i have done is i have eliminated $x$ from both the equations getting:
$$(4-y-z)^2+y^2=9$$
$$z^2+y^2+16-8y+2yz-8z+y^2=9$$ $$2y^2+2yz-8y-8z+z^2+7=0$$
But if i eliminate $y$ i get a different equation?
A cylinder and a plane do not intersect in a surface, but in a curve $\gamma$. This $\gamma$ (an ellipse) is a one-dimensional object in ${\mathbb R}^3$. Therefore it cannot be described by a single equation in the three variables $x$, $y$, $z$.
If you allow two equations then you can as well take the two equations $x^2+y^2=9$ and $x-y+z=4$ given at the outset.
Another thing is a parametrization of the ellipse. Here we use the polar angle $\phi$ as parameter. The coordinates $x$ and $y$ then just describe the given cylinder boundary, and $z$ is computed from the plane equation. In this way we obtain the parametrization $$\gamma:\quad\phi\mapsto{\bf r}(\phi):=\bigl(3\cos\phi, \>3\sin\phi, \>4-3\cos\phi+3\sin\phi\bigr)\qquad(0\leq\phi\leq2\pi)\ .$$