SOLUTION: "Find an equation of the locus of all points such that the difference of their distances from (6,0) and (-6,0) is always equal to 2."

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Question 824164: "Find an equation of the locus of all points such that the difference of their distances from (6,0) and (-6,0) is always equal to 2."
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
"Find an equation of the locus of all points such that the difference of their distances from (6,0) and (-6,0) is always equal to 2." tells us:
  • ...that the locus is a hyperbola
  • ...that the two points from whom the difference in the distances is 2 are the foci of the hyperbola
  • ...that the hyperbola has a horizontal transverse axis since the foci are on the same horizontal line (y = 0)
  • ...that the center is (0, 0) since the center is always halfway between the two foci
  • ...that the value for "c" for this hyperbola is 6 since the distance from the center to a focus is "c"
  • ...that the value for "a" is 1 since the constant difference, 2, is always equal to 2a
From all this we are almost ready to write the equation. The general equation for a horizontal hyperbola is:
%28x-h%29%5E2%2Fa%5E2+-+%28y-k%29%5E2%2Fb%5E2+=+1
where "h" and "k" are the coordinates of the center. Since we already have the center and the "a", all we need is the "b".

To find "b" we use the fact the for all hyperbolas
c%5E2+=+a%5E2+%2B+b%5E2
Substituting in the know values for "a" and "c":
%286%29%5E2+=+%281%29%5E2+%2B+b%5E2
And we solve for b (or b%5E2 since that is what we need for the equation. Simplifying...
36+=+1+%2B+b%5E2
Subtracting 1:
35+=+b%5E2

We're now ready for the equation. Substituting in the values for h, k, a and b%5E2 into
%28x-h%29%5E2%2Fa%5E2+-+%28y-k%29%5E2%2Fb%5E2+=+1
we get:
%28x-%280%29%29%5E2%2F%281%29%5E2+-+%28y-%280%29%29%5E2%2F%2835%29+=+1
Simplifying...
x%5E2%2F%281%29%5E2+-+y%5E2%2F%2835%29+=+1
x%5E2%2F1+-+y%5E2%2F%2835%29+=+1
This is the desired equation.

P.S. Although the first term simplifies down to just x%5E2 we usually leave the terms in fraction form so that the equation more closely resembles the general form.