What does "c" represent in this equation? $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=c$ In the following equation of an ellipse what does "c" stand for? I first thought it is just some scale factor but I am not sure. I know that normally the value used is 1 but my problem comes from me believing that "c" is a scale factor. 

I believe "c" is a scale factor yet I do not understand how it works, let me illustrate: 

In this diagram I made using Desmos both ellipses are exactly the same yet the red one was made using draw.io (I made it have the same length and width as the Desmos-made ellipse yet when I screenshot it and moved it onto Desmos it was resized) and the blue one using Desmos (as evidenced on the left side of the image). Now clearly this draw.io ellipse underwent a maginification of 0.5x since its real dimension are 12 units by 8 units, yet it crosses the x-axis at (-6,0) and (6,0) and the y-axis at (0,-4) and (0,4). Nonetheless when I take this into consideration in the Desmos-made equation the ellipse does not resize accordingly. 

Can someone explain to me why this happens? "c" in this equation should equal .25.
Edit: I made a mistake when constructing my question. What I really wanted to ask was how do I find "c" given the ellipse made in draw.io?
 A: $\sqrt{c}a$ and $\sqrt{c}b$ are the semimajor and semiminor axes of the ellipse.
A: $$\begin{align}
\frac{x^2}{6^2}+\frac{y^2}{4^2}=\frac12 \\[2ex]
\frac{x^2}{1/2\times6^2}+\frac{y^2}{1/2\times4^2}=\frac{1/2}{1/2} \\[2ex]
\frac{x^2}{6^2/2}+\frac{y^2}{4^2/2}=1 \\[2ex]
\frac{x^2}{(6/\sqrt2)^2}+\frac{y^2}{(4/\sqrt2)^2}=1 \\[2ex]
\end{align}$$
Thus $c$ scales the radii. Switching $1$ for $c$ scales the radii by a factor of $\sqrt c$.
A: You have one degree of redundancy in the definition.  The usual definition of an ellipse centered at the origin (so $h=k=0$, which is not important here) is $$\frac {x^2}{a^2}+\frac {y^2}{b^2}=1$$  Your equation is the same if you divide both sides by $c$ (as long as $c \gt 0$, otherwise you have a hyperbola).  This changes the usual $a$ to $\frac a{\sqrt c}$ and $b$ similarly.  We usually try to make the standard equation such that no constant is redundant, which is why we use the definition I quoted.
The answer to your added question is that you can't.  As I showed, $c$ is redundant.  You can determine $\frac {a^2}c$, but not $a$ and $c$ independently in your equation.
