Finding the centre of the ellipse $(X')^2 + 6(Z')^2 - 9Z' + 3 = 0$

Question

I am told that

$$(X')^2 + 6(Z')^2 - 9Z' + 3 = 0$$

is an ellipse, and now I'm trying to find the centre of this ellipse.

WolframAlpha verifies that it is indeed an ellipse, but I'm struggling to see how it is. The general equation of an ellipse with center $(h, k)$ and major axis parallel to the $x$-axis is

$$\dfrac{(x - h)^2}{a^2} + \dfrac{(y - k)^2}{b^2} = 1$$

And the general equation of an ellipse with center $(h, k)$ and major axis parallel to the $y$-axis is

$$\dfrac{(x - h)^2}{b^2} + \dfrac{(y - k)^2}{a^2} = 1$$

However, the equation of the supposed ellipse does not resemble any of these, and nor am I able to manipulate it so that it does.

So how is $(X')^2 + 6(Z')^2 - 9Z' + 3 = 0$ an ellipse? And how do I find its centre?

I would greatly appreciate it if people would please take the time to clarify this.

Answer 1

$(X')^2+6(Z')^2-9Z'+3=0$ is the same as $(X')^2+6\left(Z'-\frac34\right)^2-\frac{54}{16}+3=0$

or $(X')^2+6\left(Z'-\frac34\right)^2=\frac38$.

Answer 2

If you're interested in just finding the coordinates of the center, then it's point where the gradient of the quadratic vanishes. The gradient is $(2x,12z-9)$. Thus, $$2x=0\\12z-9=0 $$ and we see it's $(0,3/4)$.