document.write( "Question 1142079: Find the coordinate of the vertices and foci and
\n" ); document.write( "the equation of the asymptotes for the hyperbola
\n" ); document.write( "with the equation (x+6)^2/36-(y+3)^2/9=1 \n" ); document.write( "
Algebra.Com's Answer #762794 by greenestamps(13200)\"\" \"About 
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\n" ); document.write( "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

\n" ); document.write( "\"%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2+=+1\"

\n" ); document.write( "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.

\n" ); document.write( "With that much information, you can find the coordinates of the center and the two vertices.

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

\n" ); document.write( "Now you can find the coordinates of the two foci.

\n" ); document.write( "For the asymptotes, set the equation equal to 0 instead of 1 and solve.

\n" ); document.write( "\"%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2+=+0\"

\n" ); document.write( "The expression on the left is a difference of squares, so it is easy to solve.

\n" ); document.write( "\"%28%28x-h%29%2Fa-%28y-k%29%2Fb%29%28%28x-h%29%2Fa%2B%28y-k%29%2Fb%29+=+0\"

\n" ); document.write( "The equations of the two asymptotes come from solving the equations

\n" ); document.write( "\"%28x-h%29%2Fa-%28y-k%29%2Fb+=+0\" and \"%28x-h%29%2Fa%2B%28y-k%29%2Fb%29=+0\" \n" ); document.write( "
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