A curve is traced by a point P (x,y) which moves such that its distance from the point A(-1,1) is three times its distance from the point B(2,-1). Determine the equation of the curve.
2007-07-11 16:54:07 · 4 answers · asked by lets.dancex3 1 in Science & Mathematics ➔ Mathematics
(x+1)^2+(y-1)^2=(x-2)^2+(y+1)^2
x^2+y^2+2x-2y+2=x^2+y^2-4x+2y+5
2x-2y+2=-4x+2y+5
6x-4y-3=0
line equation
2007-07-11 17:06:00 · answer #1 · answered by iyiogrenci 6 · 0⤊ 1⤋
sqrt[(y-1)^2 + (x+1)^2] = 3 sqrt[(y+1)^2 + (x-2)^2]
Square both sides:
[(y-1)^2 + (x+1)^2] = 9 [(y+1)^2 + (x-2)^2]
[(y-1)^2 + (x+1)^2] = 9(y+1)^2 + 9(x-2)^2
Expand:
y^2 - 2y + 1 + x^2 + 2x + 1 = 9y^2 + 18y + 9 + 9x^2 - 36x + 36
Combine terms:
8y^2 + 20y + 8 + 8x^2 - 38x + 35 = 0
8y^2 + 20y + 8x^2 - 38x = - 43
Complete the square:
8(y^2 + 2.5y + ___) + 8(x^2 - 4.75x + ___) = - 43
8(y^2 + 2.5y + (2.5/2)^2) + 8(x^2 - 3.4x + (-4.75/2)^2) = -43 + 12.5 + 45.125
8(y + 1.25)^2 + 8(x - 2.375)^2 = 14.625
Divide both sides by 8:
(y + 1.25)^2 + (x - 2.375)^2 = 1.83
2007-07-12 00:10:06 · answer #2 · answered by whitesox09 7 · 0⤊ 0⤋
My Brain Hurts
2007-07-12 00:01:12 · answer #3 · answered by Anonymous · 0⤊ 0⤋
we have something in common...i dont get it either.
2007-07-11 23:57:24 · answer #4 · answered by Linda 1 · 0⤊ 0⤋