Hello, bob1!
This requires the Distance Formula and a
lot of careful algebra . . .
Use the definition of an ellipse to find an equation of the ellipse
having foci \(\displaystyle F_1(1,\,1)\) and \(\displaystyle F_2(-1,\,-1)\) and sum of focal radii 3.
The answer to this problem is: \(\displaystyle 20x^2\,+\,20y^2\,-\,32xy\,-\,9\:=\:0\;\)
Let \(\displaystyle P(x,y)\) be a point on the ellipse . . . then: \(\displaystyle \,\overline{PF}_1\,+\,\overline{PF}_2\:=\:3\)
Then we have: \(\displaystyle \,\sqrt{(x\,-\,1)^2\,+\,(y\,-\,1)^2}\.+\,\sqrt{(x\,+\,1)^2\,+\,(y\,+\,1)^2}\;=\;3\)
Isolate a radical: \(\displaystyle \,\sqrt{(x\,-\,1)^2\,+\,(y\,-\,1)^2}\;=\;3\,-\,\sqrt{(x\,+\,1)^2\;+\;(y\,+\,1)^2}\)
Square both sides: \(\displaystyle \,(x\,-\,1)^2\,+\,(y\,-\,1)^2\;=\;9\,-\,6\sqrt{(x\,+\,1)^2\,+\,(y\,+\,1)^2}\,+\,(x\,+\,1)^2\,+\,(y\,+\,1)^2\)
Expand: \(\displaystyle \,\not{x^2}\,-\,2x\,+\,\not{1}\,+\,\not{y^2}\,-\,2y\,+\,\not{1}\;=\;9\,-\,6\sqrt{(x\,+\,1)^2\,+\,(y\,+\,1)^2}\,+\,\not{x^2}\,+\,2x\,+\,\not{1}\,+\,\not{y^2}\,+\,2y\,+\,\not{1}\)
Simplify: \(\displaystyle \,6\sqrt{(x\,+\,1)^2\,+\,(y\,+\,1)^2} \;= \;4(x\,+\,y)\,+\,9\)
Square both sides: \(\displaystyle \,36\left[(x\,+\,1)^2\,+\,(y\,+\,1)^2\right]\;=\;16(x\,+\,y)^2\,+\,72(x\,+\,y)\,+\,81\)
Expand: \(\displaystyle \,36x^2\,+\,\sout{72x}\,+\,36\,+\,36y^2\,+\,\sout{72y}\,+\,36\;=\;16x^2\,+\,32xy\,+\,16y^2\,+\,\sout{72x}\,+\,\sout{72y}\,+\,81\)
Therefore: \(\displaystyle \:20x^2\,+\,20y^2\,-\,32xy\,-\,9\;=\;0\;\;\)
. . . ta-DAA!