Parametric equation of a cone

Question

I usually use the following parametric equation to find the surface area of a regular cone $z=\sqrt{x^2+y^2}$: $$x=r\cos\theta$$ $$y=r\sin\theta$$ $$z=r$$

And make $0\leq r \leq 2\pi$, $0 \leq \theta \leq 2\pi$.

I've now have a cone $z=\sqrt{2x^2+2y^2}$ and I think the parametric equation I normally use won't work anymore. Which would be a more suitable one in this case? Is there any generic parametric equation for cones, because one of the form $z=\sqrt{4x^2+y^2}$ would also have a different one.

Answer 1

Let $z=\sqrt{a^2 x^2 + b^2 y^2}$ where $a>0$ and $b >0$

Then let $z=r$, $x=\frac{r}{a} cos(\theta)$ and $y= \frac{r}{b} sin(\theta)$

Answer 2

Correct formula (from: https://mathworld.wolfram.com/Cone.html) in Mathematica Language is:

ParametricPlot3D[{r*(h - u)/h Cos[t], r*(h - u)/h Sin[t], u}, {u, 0, 
  h}, {t, 0, 2 Pi}]

where: $h$ is height of cone, $r$ is radius of cone (constants); $t, u$ are parameters.