- #1

fire9132

- 5

- 0

## Homework Statement

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.

## Homework Equations

None

## The Attempt at a Solution

Distance from Point A to Point P:

[tex] \sqrt[2]{(x+1)^{2} + (y-1)^{2}} [/tex]

Distance from Point P to Point B:

[tex] \sqrt[2]{(x-2)^{2} + (y+1)^{2}} [/tex]

Distance from Point A to Point P is three times the distance from Point P to Point B so...

[tex] 3 \sqrt[2]{(x+1)^{2} + (y-1)^{2}} = \sqrt[2]{(x-2)^{2} + (y+1)^{2}} \\

9(x+1)^{2} + 9(y-1)^{2} = (x-2)^{2} + (y+1)^{2} \\

9x^{2} + 18x + 9 + 9y^{2} - 18y - 9 = x^{2} - 4x + 4 + y^{2} + 2y + 2 \\

8x^{2} + 22x + 8y^{2} - 20y + 13 = 0 [/tex]

Doing this gives me the equation of a circle, which I don't think is a curve. After figuring out that the center of that circle was (-11/8, 5/4), the distance from the center to B is not 3 times the distance from the center to A. Then, I think my answer is wrong.

Reanalyzing the problem, I thought of a different approach which was to solve for the equation of a parabola knowing the directrix would be a line going through A(-1,1) and then the focus being (2,-1). However, this would make a slanted parabola and I have no idea how to make an equation for that.