# Finding a parameterization of the paraboloid $900z = 25x^2 + 36y^2$

Question:

Find a parameterization of the paraboloid $$900z = 25x^2 + 36y^2$$.

My Work

$$25x^2 + 36y^2 = 900z$$

$$\implies (5x)^2 + (6y)^2 = (30\sqrt{z})^2$$

Can we can represent this equation using cylindrical coordinates?

$$(x,y,z) = (\rho \cos(\theta), \rho \sin(\theta), \zeta)$$ where $$\rho =$$ radius and $$2\pi \ge \theta \ge 0$$.

I think I'm on the right track here, but I've spent hours unsuccessfully pondering over this problem and doing research. I want to get it into cylindrical coordinates and then parameterise it in terms of $$u$$ and $$v$$, but I'm just completely stuck.

I would greatly appreciate it if people could please take the time to show me the correct reasoning and solution for this problem.

• Your parametrization $(x,y) = (\rho \cos(\theta), \rho \sin(\theta))$ cannot work because it would mean that the level curves are circles, whereas they are ellipses. – Jean Marie Apr 9 '17 at 10:30

Whenever you have an equation of the form $z = f(x,y)$, you can use $x$ and $y$ as parameters. So one possible set of parametric equations is $$x = u \quad ; \quad y = v \quad ; \quad z = \frac{1}{900}(25u^2 + 36v^2)$$ If you really want to use trig functions, then: $$x = \tfrac15 r \cos \theta \quad ; \quad y = \tfrac16 r \sin \theta \quad ; \quad z = \tfrac{1}{900}r^2$$ True polar coordinates are going to be messy. The problem is that each curve of the form $z = \text{constant}$ is an ellipse, and the equation of an ellipse in polar coordinates is a bit complicated.