Yahoo Answers 4-15-14 This is a lot of messy algebra. If you recognize it, this will form an ellipse. Just getting there is... well... PA=√((x-(-5))^2+y^2)=√((x+5)^2+y^2 PB=√((x-5)^2+y^2) PA+PB=12 PA=-PB+12 (square both sides) PA^2=PB^2-24PB+144 24PB=PB^2-PA^2+144 let's simplify the right side PB^2=(x-5)^2+y^2=x^2-10x+25+y^2 PA^2=(x+5)^2+y^2=x^2+10x+25+y^2 PB^2-PA^2+144=-20x+144 24PB=-20x+144 (divide by 4) 6PB=-5x+36 (square both sides again) 36(x-5)^2+36y^2=25x^2-360x+1296 36x^2-360x+900+36y^2=25x^2-360x+1296 11x^2+36y^2=396 the 2 points are the foci of the ellipse x^2/36+y^2/11=1
Move point C around to see that the sum of the distances is always 12