Locus of points with distance sum =8

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The discussion focuses on finding the locus of points where the sum of distances from points A(0,-2) and B(0,2) equals 8. Participants clarify that the correct approach involves using the distance formula for both points and setting their sum equal to 8, leading to the equation √[x² + (y+2)²] + √[x² + (y-2)²] = 8. There is confusion about the concept of the locus and the need to derive an equation for it. The conversation emphasizes that this locus represents multiple points in the plane that satisfy the distance condition, ultimately forming an ellipse. The discussion concludes with a suggestion that the derived equation can be simplified to represent the locus effectively.
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Locus of points such that the sum of whose distances from A(0,-2) and B(0,2) is 8

I plotted the points its a straight line on the y axis. I sub both points into the distance formula (seperate distance formulas) and then made the eqn = 8 I squared both sides getting rid of the square root and then simplified and got
2x^2 +2y^2+8=64 I think I did this question wrong please help me :cry:
 
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their talking about the sum of the distances from A and B. So you need to come up with an equation for the the distance from A and the distance from B seperatly, and then add them...

that sum should equal 8

to see the motivation behind this problem look up the definition of an ellpise, they even give a method to solve this problem here.
http://mathworld.wolfram.com/Ellipse.html
 
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Here they are separately

Point A = square root[(x^2)+(y+2)^2]

Point B= square root [(x^2)+(y-2)^2]

I did make two different equations but when you say for the distance from A and from B I don't get FROM WHAT?

I added the above two equations and set it =8 and then solved but I think I did something wrong.

shouldnt I be finding an equation for the locus why am I verifying that point A's distance + point B's =8?
 
aisha said:
I did make two different equations but when you say for the distance from A and from B I don't get FROM WHAT?

The distance from A and from B TO any point that satisfies the conditions you were given! i.e. any point (call it C, for example) in the plane such that if you add how far C is from A, to how far C is from B, you get 8. Stated another way: the sum of the distances to point C from point A and from point B is 8. There is more than one such point 'C'...indeed they trace out a curve, and this set of points on the curve that satisfies that condition is called the locus.

Edit: the animation on that link MathStudent gave you illustrates it very well.
 
Yes I see what you mean the animation is good as well but it is hard to find the equation for the locus did you see how many steps their are? Isnt there an easier way?
 
Sure. They just simplified the equation to a nice form. The equation

\sqrt{x^2 + (y+2)^2} + \sqrt{x^2 + (y-2)^2} = 8

defines the locus just as well as the standard ellipse equation.
 
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