SOLUTION: 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.

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.
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
Standard equation for an ellipse,
(x-h)^2/a^2+(y-k)^2/b^2 = 1, a>b
(h,k) being the (x,y) coordinates of the center of the ellipse
from the given ellipse equation,
(x-2)^2/81=(y+3)^2/49 = 1,
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
a^2=81
a=9
b^2=49
b=7
c = sqrt( a^2-b^2)=sqrt(81-49)=sqrt(32) =5.66
finding focus on the left side,
starting from the ctr at x=2, move a(5.66) units to the left to get (2-5.66) =-3.66
the y-coordinate of -3 does not change
so, this is how you got (-3.66,-3) for the coordinates of the left focus
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
You can find the vertices in a similar manner using a instead of c

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