SOLUTION: Find the coordinate of the vertices and foci and the equation of the asymptotes for the hyperbola with the equation (x+6)^2/36-(y+3)^2/9=1

Algebra ->  Finance -> SOLUTION: Find the coordinate of the vertices and foci and the equation of the asymptotes for the hyperbola with the equation (x+6)^2/36-(y+3)^2/9=1      Log On


   

Question 1142079: Find the coordinate of the vertices and foci and
the equation of the asymptotes for the hyperbola
with the equation (x+6)^2/36-(y+3)^2/9=1
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The x^2 term is positive, so the branches open right and left. The general form of the equation for that kind of hyperbola is

%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2+=+1

In that form, the center is (h,k); a is the distance (in the x direction) from the center to each vertex -- i.e., from the center to each end of the transverse axis; and b is the distance (in the y direction) from the center to each end of the conjugate axis.

With that much information, you can find the coordinates of the center and the two vertices.

The distance (in the x direction) from the center to each focus is c, where c%5E2+=+a%5E2%2Bb%5E2.

Now you can find the coordinates of the two foci.

For the asymptotes, set the equation equal to 0 instead of 1 and solve.

%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2+=+0

The expression on the left is a difference of squares, so it is easy to solve.

%28%28x-h%29%2Fa-%28y-k%29%2Fb%29%28%28x-h%29%2Fa%2B%28y-k%29%2Fb%29+=+0

The equations of the two asymptotes come from solving the equations

%28x-h%29%2Fa-%28y-k%29%2Fb+=+0 and %28x-h%29%2Fa%2B%28y-k%29%2Fb%29=+0