Question

*calculus 3* Let S be the surface whose equation in rectangular coordinates is: x^2-y^2+z^2-4x-2y-2z+4=0
specify the vertex, axis of symmetry, and traces in the xy,xz, and yz-planes.

Note: I have a general idea of how to solves the traces for the planes, but I don't really know how to find the vertices or the axis of symmetry.

Answer 1

First try to find out what the surface might represent. You can easily write the given equation as a linear combination of quadratic factors by completing the square:
\[\begin{align*}
x^2-y^2+z^2-4x-2y-2z+4&=0\\\\
x^2-4x\color{red}{+4}-(y^2+2y\color{red}{+1})+z^2-2z\color{red}{+1}+4&=\color{red}{4-1+1}\\\\
(x-2)^2-(y+1)^2+(z-1)^2&=0\\\\
y&=-1\pm\sqrt{(x-2)^2+(z-1)^2}
\end{align*}\]
This is the equation for a double cone (like the one in the figure here: http://en.wikipedia.org/wiki/Cone#Other_mathematical_meanings )

The vertex of the cone is where the two cones meet. In this case, it's at the point where both \(x=z=0\), which gives \(y=-1\), and hence the point \((0,-1,0)\).

Answer 2

Ah got it! Thanks, i was having trouble finding out what the vertex was for the shape!