Curve is traced by a point P(x,y)?

Question

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 poinit B(2,-1). Determine the equation of the curve.

Answer 1

It sounds like you have something like

9((x + 1)^2 + (y-1)^2) = (x-2)^2 + (y+1)^2

The above is the equation for the distances you specified with each side squared to eliminate the square root.

Expanding and combining terms

9(x^2 + 2x + 1 + y^2 - 2y + 1) = x^2 -4x + 4 + y^2 + 2y + 1

8x^2 + 22x + 13 + 8y^2 -4y = 0

Now for each x value, there will be a quadratic equation resulting in 2 y values. It looks like you will have squared function between these two points. If the y value is imaginary then there is no valid point to graph.

Answer 2

using the info given actually plot the acutal points on graph. Using y=mx+b determine linear equation if it's not that form just enter in ti-83 calc and under trace functions you can obtain equation.

Answer 3

what you're trying to do is set up the equation for an ellipse...

just use the distance formula and set "distance to A" = 3 times "distance to B"

Simplify it to get your equation

A and B will turn out to be the foci of the ellipse

Answer 4

hmm.. i guess u'd just simplify this equation

(x+1)^2 + (y-1)^2 = 3 [ (x-2)^2 + (y+1)^2 ]