Determining the equation of a curve.

  • Thread starter Thread starter fire9132
  • Start date Start date
  • Tags Tags
    Curve
AI Thread Summary
The discussion centers on determining the equation of a curve traced by a point P(x,y) based on its distances from two fixed points A(-1,1) and B(2,-1). The initial approach involved setting the distance from A to P as three times that from P to B, leading to an equation that was mistakenly interpreted as a circle. After realizing the error, the user considered a parabolic approach, identifying A as the directrix and B as the focus. Feedback from other participants highlighted a misplacement of the factor of three in the equation, prompting a re-evaluation of the calculations. The conversation emphasizes the importance of careful algebraic manipulation and understanding the geometric properties involved.
fire9132
Messages
5
Reaction score
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:
\sqrt[2]{(x+1)^{2} + (y-1)^{2}}
Distance from Point P to Point B:
\sqrt[2]{(x-2)^{2} + (y+1)^{2}}

Distance from Point A to Point P is three times the distance from Point P to Point B so...
3 \sqrt[2]{(x+1)^{2} + (y-1)^{2}} = \sqrt[2]{(x-2)^{2} + (y+1)^{2}} \\<br /> 9(x+1)^{2} + 9(y-1)^{2} = (x-2)^{2} + (y+1)^{2} \\<br /> 9x^{2} + 18x + 9 + 9y^{2} - 18y - 9 = x^{2} - 4x + 4 + y^{2} + 2y + 2 \\<br /> 8x^{2} + 22x + 8y^{2} - 20y + 13 = 0

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.
 
Physics news on Phys.org
Arcs and circles are curves. A curve describes any figure which is not a straight line.
 
fire9132 said:

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:
\sqrt[2]{(x+1)^{2} + (y-1)^{2}}
Distance from Point P to Point B:
\sqrt[2]{(x-2)^{2} + (y+1)^{2}}

Distance from Point A to Point P is three times the distance from Point P to Point B so...
3 \sqrt[2]{(x+1)^{2} + (y-1)^{2}} = \sqrt[2]{(x-2)^{2} + (y+1)^{2}} \\<br /> 9(x+1)^{2} + 9(y-1)^{2} = (x-2)^{2} + (y+1)^{2} \\<br /> 9x^{2} + 18x + 9 + 9y^{2} - 18y - 9 = x^{2} - 4x + 4 + y^{2} + 2y + 2 \\<br /> 8x^{2} + 22x + 8y^{2} - 20y + 13 = 0

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.
 
  • Like
Likes 1 person
fire9132 said:
3 \sqrt[2]{(x+1)^{2} + (y-1)^{2}} = \sqrt[2]{(x-2)^{2} + (y+1)^{2}}
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!
 
  • Like
Likes 1 person
Wow, I feel stupid for doing that. Finally got it now. Thank you all!
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top