How to sketch this quadric surface

Question

I figured out that this quadric surface is one sheeted hyperboloid but I am unable to plot it in the $xy, xz,$ and the $yz$- planes. I have used the approach where we can put the one axis zero and find the points at that plane, but it's not working out for me. I would really appreciate it if someone could please share a detailed solution. Thank you!

Answer 1

It can be tricky to visualize three-dimensional surfaces from the equation alone, but plotting in the plane isn't so bad when remember that (e.g.) the XY plane is the set of all points $(x,y,z)$ where $z=0$. This means that on the XY plane, our equation $$ z^2 = 4x^2 + y^2 + 8x − 2y + 4z $$ simplifies by setting $z=0$ to $$ 0 = 4x^2 + y^2 + 8x - 2y. $$ If you have a hard time seeing what shape this should make, you may wish follow the advice in the comments by completing the square and factoring: $$ 0 = 4(x^2 + 2x + 1 - 1) + (y^2 - 2y + 1 - 1) = 4(x + 1)^2 + (y - 1)^2 - 5. $$ Rearrange a teeny bit, and we get $$ 5 = 4(x + 1)^2 + (y - 1)^2. $$ Does this look familiar?

Similarly, the XZ plane is the set of all points $(x,y,z)$ where $y=0$, so we want to plot $$ z^2 = 4x^2 + 8x + 4z. $$ Do you see where to go from here? And the same for the YZ plane?

If you get stuck, might I recommend https://www.desmos.com/calculator for checking your two-dimensional graphs. You can also, for example, input the complete three-dimensional equation and show XY slices as a function of $z$.

Answer 2

Here’s how I would attack the problem, but you must check my computations:

\begin{align} z^2&=4x^2 + y^2 + 8x -2y + 4z\\ z^2-4z&=4(x^2+2x)\quad+\quad(y^2-2y)\\ (z-2)^2-4&=4(x+1)^2-4\quad+\quad(y-1)^2-1\\ \zeta^2&=4\xi^2+\eta^2-1\,, \end{align} and I guess that says that you have a hyperboloid of one sheet, center at $(-1,1,2)$.

(I notice belatedly that the first commenter, @Gio, has given you this answer already.)