document.write( "Question 390231: One focus is at (-3.66,-3).Find the other focus for the ellipse defined by this equation (x-2)^2/81 + (y+3)^2/49=1. I have no Idea how to do this please help,I'd appreciate it. \n" ); document.write( "
Algebra.Com's Answer #276750 by lwsshak3(11628)\"\" \"About 
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Standard equation for an ellipse,
\n" ); document.write( "(x-h)^2/a^2+(y-k)^2/b^2 = 1, a>b
\n" ); document.write( "(h,k) being the (x,y) coordinates of the center of the ellipse
\n" ); document.write( "from the given ellipse equation,
\n" ); document.write( "(x-2)^2/81=(y+3)^2/49 = 1,
\n" ); document.write( "by inspection, it is seen that the ctr is at (2,-3), and the ellipse takes a horizontal shape because a^2 is the denominator of (x-2)^2. It would take a vertical shape if a^2 was the denominator of (y+3)^2
\n" ); document.write( "a^2=81
\n" ); document.write( "a=9
\n" ); document.write( "b^2=49
\n" ); document.write( "b=7
\n" ); document.write( "c = sqrt( a^2-b^2)=sqrt(81-49)=sqrt(32) =5.66
\n" ); document.write( "finding focus on the left side,
\n" ); document.write( "starting from the ctr at x=2, move a(5.66) units to the left to get (2-5.66) =-3.66
\n" ); document.write( "the y-coordinate of -3 does not change
\n" ); document.write( "so, this is how you got (-3.66,-3) for the coordinates of the left focus
\n" ); document.write( "similarly, for the right focus, start from x=2 and move 5.66 units to the right to get (7.66,-3) for the coordinates of the right focus\r
\n" ); document.write( "\n" ); document.write( "You can find the vertices in a similar manner using a instead of c
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