Determining the equation of a curve.

  • Thread starter fire9132
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  • #1
fire9132
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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.
 

Answers and Replies

  • #2
SteamKing
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Arcs and circles are curves. A curve describes any figure which is not a straight line.
 
  • #3
Ray Vickson
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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.

You have written 3*d(A,P) = d(P,B), the exact opposite of what you want.
 
  • #4
skiller
237
5
[tex] 3 \sqrt[2]{(x+1)^{2} + (y-1)^{2}} = \sqrt[2]{(x-2)^{2} + (y+1)^{2}}[/tex]
You have placed the multiple 3 on the wrong side of the equation here. Try again from that point.

Also, be careful when you expand the expressions as you've made a couple of sloppy errors in your subsequent lines of working too.

EDIT: Beaten to it by Ray!
 
  • #5
fire9132
5
0
Wow, I feel stupid for doing that. Finally got it now. Thank you all!
 

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