Hello, k9fireman!
Trying to understand how: \(\displaystyle \sqrt{(x+c)^2\,+\,y^2}\,+\,\sqrt{(x-c)^2\,+\,y^2}\;=\;2a\)
simplifies to: \(\displaystyle \,(a^2\,-\,c^2)x^2\,+\,a^2y^2\;=\;a^2(a^2\,-\,c^2)\)
First, "isolate" a radical: \(\displaystyle \,\sqrt{(x+c)^2\,+\,y^2}\;=\;2a\,-\,\sqrt{x-c)^2\,+\,y^2}\)
Square both sides: \(\displaystyle \,(x+c)^2\,+\,\not{y^2}\;=\;4a^2\,-\,4a\sqrt{(x-c)^2\,+\,y^2}\,+\,(x-c)^2\,+\,\not{y^2}\)
We have: \(\displaystyle \,\not{x^2}\,+\,2cx\,+\,\not{c^2}\,+\,\not{y^2}\;=\;4a^2\,-\,4a\sqrt{(x-c)^2\,+\,y^2}\,+\,\not{x^2}\,-2cx\,+\,\not{c^2}\,+\,\not{y^2}\)
Then: \(\displaystyle \,4a\sqrt{(x-c)^2\,+\,y^2}\;=\;4a^2\,-\,4cx\;\;\Rightarrow\;\;a\sqrt{(x-c)^2\,+\,y^2}\;=\;a^2\,-\,cx\)
Square both sides: \(\displaystyle \,a^2[(x-c)^2\,+\,y^2]\;=\;a^4\,-\,2a^2cx\,+\,c^2x^2\)
Expand: \(\displaystyle \,a^2x^2\,-\,\sout{2a^2cx}\,+\,a^2c^2\,+\,a^2y^2\;=\;a^4\,-\,\sout{2a^2cx}\,+\,c^2x^2\)
Then: \(\displaystyle \,a^2x^2\,-\,c^2x^2\,+\,a^2y^2\;=\;a^4\,-\,a^2c^2\)
And: \(\displaystyle \,(a^2\,-\,c^2)x^2\,+\,a^2y^2\;=\;a^2(a^2\,-\,c^2)\;\;\)
. . . There!
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Divide by \(\displaystyle a^2(a^2\,-\,c^2):\L\;\;\frac{x^2}{a^2}\,+\,\frac{y^2}{a^2\,-\,c^2}\;=\;1\)
Let: \(\displaystyle b^2\:=\:a^2\,-\,c^2\) . . . and we have: \(\displaystyle \L\,\frac{x^2}{a^2} \,+ \,\frac{y^2}{b^2}\;=\;1\;\;\)
. . . ta-DAA!