Eliminating $m$ from the system $x=\frac{\sqrt{a^2-b^2}+\sqrt{a^2+b^2m^2}}{\sqrt{1+m^2}}$, $y=\frac{m\sqrt{a^2-b^2}+\sqrt{a^2m^2+b^2}}{\sqrt{1+m^2}}$ I have 2 equations and need to eliminate the variable $m$ from them, however the equations quickly get drastically complex:
$$x = \frac{\sqrt{a^2-b^2}+\sqrt{a^2+b^2m^2}}{\sqrt{1+m^2}}$$
$$y = \frac{m\sqrt{a^2-b^2}+\sqrt{a^2m^2+b^2}}{\sqrt{1+m^2}}$$
I have wasted hours on my written approach and also tried many online equations solving websites, but they are no good to me.
In case when $a=2$ and $b=1$ the following relation should be obtained:
$$x^2 + y^2 + \frac{1}{x^2} + \frac{1}{y^2} = 16$$
Just to make sure I calculated the above quantity using desmos and changing the variable $m$, keeping $a$ and $b$ as $2$ and $1$ respectively does indeed had no effect on it.
Please help me out or just let me know if finding a general solution is really possible here, any kind of help will be greatly appreciated.
Here are some extra details, where this question actually comes from:
I was trying to find the locus of focus of an ellipse which rotates in the first quadrant such that it always touches the coordinate axes.
To start with, I wrote an equation of ellipse having it's major and minor axes as $mx-y+c=0$ and $x+my+d=0$  respectively, applying the given condition gives
$$c = \frac{-m\sqrt{a^2+b^2m^2} + \sqrt{b^2+a^2m^2}}{\sqrt{1+m^2}}$$
$$d = -\frac{m\sqrt{b^2+a^2m^2} + \sqrt{a^2+b^2m^2}}{\sqrt{1+m^2}}$$
Now the above given value of $x,y$ is actually the coordinates of one of it's foci, which we can find by shifting the coordinate axes to given major and minor axes, from which eliminating $m$ should give the desired locus.

$$x^2+y^2+\frac{b^4}{x^2}+\frac{b^4}{y^2}=4a^2$$
Let $A=a^2m^2+b^2$, $B=a^2+b^2m^2$, $M=m^2+1$ and $D=a^2-b^2$. Then
$$x=\sqrt{\frac BM}+\sqrt{\frac DM}\\
\frac{b^2}x=\sqrt{\frac BM}-\sqrt{\frac DM} \\
x-\frac{b^2}x=2\sqrt{\frac DM}\\
y-\frac{b^2}y=2m\sqrt{\frac DM}\\
\left(x-\frac{b^2}x\right)^2+\left(y-\frac{b^2}y\right)^2=4D$$
