I might be a bit slow on math, but I haven't seen this explicitly described anywhere, I'm reading a paper and in the paper it has the equation: $$\Delta d=\sqrt{y^2+(x+D/2)^2}-\sqrt{y^2+(x-D/2)^2}$$ After which it says
After some simplifications, this equation can be rewritten in a more compact form as $$\frac{x^2}{a}-\frac{y^2}{b}=\frac{x^2}{\Delta d^2/4}-\frac{y^2}{D^2/4-\Delta d^2/4}=1$$
How does it go from the first equation to the second one?
(the paper equation 4 and 5)
On
$$\Delta d=\sqrt{y^2+(x+D/2)^2}-\sqrt{y^2+(x-D/2)^2}$$$$\therefore\sqrt{y^2+(x+D/2)^2}=\Delta d+\sqrt{y^2+(x-D/2)^2}$$$$\therefore y^2+(x+D/2)^2=\left(\Delta d+\sqrt{y^2+(x-D/2)^2}\right)^2$$$$\therefore y^2+x^2+Dx+D^2/4=\Delta d^2+y^2+(x-D/2)^2+2\Delta d\sqrt{y^2+(x-D/2)^2}$$$$=\Delta d^2+y^2+x^2-Dx+D^2/4+2\Delta d\sqrt{y^2+(x-D/2)^2}$$$$\therefore2\Delta d\sqrt{y^2+(x-D/2)^2}=2Dx-\Delta d^2$$$$\therefore4\Delta d^2(y^2+(x-D/2)^2)=4D^2x^2-4Dx\Delta d^2+\Delta d^4$$$$\therefore4\Delta d^2(y^2+x^2-Dx+D^2/4)=4D^2x^2-4Dx\Delta d^2+\Delta d^4$$$$\therefore4\Delta d^2y^2+4\Delta d^2x^2-4\Delta d^2Dx+\Delta d^2D^2=4D^2x^2-4Dx\Delta d^2+\Delta d^4$$$$\therefore4\Delta d^2y^2+4(\Delta d^2-D^2)x^2=\Delta d^2(\Delta d^2-D^2)$$$$\therefore\frac{4y^2}{\Delta d^2-D^2}+\frac{4x^2}{\Delta d^2}=1$$Hopefully you can see how to get to their expression from here
(I'll write $\Delta d$ as $d$ here to save typing.) Let $p=y^2 + (x + D/2)^2$ and $q=y^2 + (x - D/2)^2$ then $d=\sqrt{p}-\sqrt{q}$ so $d+\sqrt{q}=\sqrt{p}$.
Square both sides to get $$d^2+2d\sqrt{q}+q=p.$$ Rearrange this to get the final root on its own: $$\sqrt{q}=\frac1{2d}(p-q-d^2).$$ Now notice that $p-q=4Dx/2$ so the right hand side of the above can be simplified to $Dx/d-d/2$. Then square both sides to get $$ y^2+(x-D/2)^2=D^2x^2/d^2-Dx+d^2/4 $$ The $Dx$ terms on either side cancel each other, leaving $$ y^2+(1-(D/d)^2)x^2=(d^2-D^2)/4. $$ From there you are on your own.
Merry Christmas.