In Euclidean three-space, we can define paraboloidal coordinates $(u,v,\phi)$ via \begin{align*} x = uv\cos\phi,\quad y = uv\sin\phi,\quad z = \frac{1}{2}(u^2-v^2) \end{align*} Are there any singular points?
This question feels a bit vague to me because I don't really know how to graph this. I know that there wouldn't be a singular point for a paraboloid of the form
\begin{align*} x^2+y^2=z \end{align*}
But seeing these three equations makes me unsure now.
Hint: Look at the Jacobian matrix $$ \begin{bmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial y}{\partial u} & \dfrac{\partial z}{\partial u}\\ \dfrac{\partial x}{\partial v} & \dfrac{\partial y}{\partial v} & \dfrac{\partial z}{\partial v}\\ \dfrac{\partial x}{\partial \phi} & \dfrac{\partial y}{\partial \phi} & \dfrac{\partial z}{\partial \phi}\\ \end{bmatrix}. $$ The singular points are the points where the determinant of the Jacobian matrix is zero.