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Proof 13: Ellipsis equation

The circle equation (see Proof 12) can be rearranged:

$$(x-c_1)^2+(y-c_2)^2=r^2\\
\frac{(x-c_1)^2}{r^2}+\frac{(y-c_2)^2}{r^2}=1\\
\left(\frac{x-c_1}{r}\right)^2+\left(\frac{y-c_2}{r} \right)^2 =1
$$

The $x-c_1$ and $y-c_2$ distances are being ‘squeezed’ by $r$, so to say. (‘Squeezed’ in the sense that we divide by $r$ and thus make the value smaller.) This ‘squeezing’ is fixed – we cannot just change $r$, because then the circle wouldn’t fit the distances $x-c_1$ and $y-c_2$ anymore.

But we can imagine ‘squeezing’ a bit more on $x-c_1$ and a bit less on $y-c_2$. As if you squeeze a hula hoop from the sides. Then you get an ellipsis. We’ll have to name the symbols differently, such as $r_1$ and $r_2$, when the squeezing is not the same. And that gives us the ellipsis equation:

$$\left(\frac{x-c_1}{r_1}\right)^2+\left(\frac{y-c_2}{r_2} \right)^2 =1\quad_\blacksquare$$

While $r$ was the radius in the circle, the $r_1$ and $r_2$ are the “radiusses” across the longest and slimmest part of the ellipsis. We’ll instead call them half-axes, since they are half the axis-length along those longest and slimmest parts.

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