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." tells us:<ul><li>...that the locus is a hyperbola</li><li>...that the two points from whom the difference in the distances is 2 are the foci of the hyperbola</li><li>...that the hyperbola has a horizontal transverse axis since the foci are on the same horizontal line (y = 0)</li><li>...that the center is (0, 0) since the center is always halfway between the two foci</li><li>...that the value for "c" for this hyperbola is 6 since the distance from the center to a focus is "c"</li><li>...that the value for "a" is 1 since the constant difference, 2, is always equal to 2a</li></ul>From all this we are almost ready to write the equation. The general equation for a horizontal hyperbola is:
{{{(x-h)^2/a^2 - (y-k)^2/b^2 = 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".<br>
To find "b" we use the fact the for all hyperbolas
{{{c^2 = a^2 + b^2}}}
Substituting in the know values for "a" and "c":
{{{(6)^2 = (1)^2 + b^2}}}
And we solve for b (or {{{b^2}}} since that is what we need for the equation. Simplifying...
{{{36 = 1 + b^2}}}
Subtracting 1:
{{{35 = b^2}}}<br>
We're now ready for the equation. Substituting in the values for h, k, a and {{{b^2}}} into
{{{(x-h)^2/a^2 - (y-k)^2/b^2 = 1}}}
we get:
{{{(x-(0))^2/(1)^2 - (y-(0))^2/(35) = 1}}}
Simplifying...
{{{x^2/(1)^2 - y^2/(35) = 1}}}
{{{x^2/1 - y^2/(35) = 1}}}
This is the desired equation.<br>
P.S. Although the first term simplifies down to just {{{x^2}}} we usually leave the terms in fraction form so that the equation more closely resembles the general form.