Yahoo Answers 4-15-14
- Author:
- Michael Brown
http://imgur.com/w33hVuz
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