SOLUTION: Find the foci of the hyperbola defined by the equation: (x-6)^2/16 - (y+6)^2/4 = 1.I keep getting this question wrong I though I knew how to do this.Please help.

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the foci of the hyperbola defined by the equation: (x-6)^2/16 - (y+6)^2/4 = 1.I keep getting this question wrong I though I knew how to do this.Please help.      Log On


   



Question 391784: Find the foci of the hyperbola defined by the equation: (x-6)^2/16 - (y+6)^2/4 = 1.I keep getting this question wrong I though I knew how to do this.Please help.
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
From the equation we can see
  • that the center of the hyberbola is (6, -6).
  • that the major axis of the hyperbola is horizontal because the x%5E2 term is positive (and the y%5E2 term is negative (or subtracted)).
  • that a%5E2+=+16 and b%5E2+=+4

"c" is the distance from the center to the foci along the major axis. So to find the foci we will need to find "c". The equation for hyperbolas that connects a, b and c is:
c%5E2+=+a%5E2+%2B+b%5E2
Using the values we already have for a%5E2 and b%5E2 this equation becomes:
c%5E2+=+16+%2B+4
which simplifies to
c%5E2+=+20
Finding the square root of each side we get:
c+=+sqrt%2820%29
which simplifies as follows:
c+=+sqrt%284%2A5%29
c+=+sqrt%284%29%2Asqrt%285%29
c+=+2%2Asqrt%285%29

So the foci are a distance of 2sqrt%285%29, horizontally and in both directions, from the center. Since we want to move hoziontally from the center, in both directions, we will both add and subtract 2sqrt%285%29 to/from the x-coordinate of the center. So the foci will be:
(6+%2B+2sqrt%285%29, -6) and (6+-+2sqrt%285%29, -6)