SOLUTION: What are the foci of the ellipse given by the equation [((x-2)^2)/36]+[((y-8)^2)/144]=1? Please explain! I am having trouble with this. Thank you!

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: What are the foci of the ellipse given by the equation [((x-2)^2)/36]+[((y-8)^2)/144]=1? Please explain! I am having trouble with this. Thank you!      Log On


   

Question 462331: What are the foci of the ellipse given by the equation [((x-2)^2)/36]+[((y-8)^2)/144]=1?
Please explain! I am having trouble with this.
Thank you!
Answer by graphmatics(170) About Me  (Show Source):
You can put this solution on YOUR website!
((x-2)^2)/36+((y-8)^2)/144=1
From the general equation
[((x-h)^2)/a^2]+[((y-k)^2)/b^2]=1 we know that the center of the ellipse is at (h,k). So for our ellipse equation the center of the ellipse is at (2,8).
c is the distance from the center to a focus point. The expression for c (when b > a) is
c+=+sqrt%28+b%5E2-a%5E2+%29+
So for our ellipse
c+=+sqrt%28+12%5E2-6%5E2+%29+
c+=+sqrt%28+144-36+%29+
c+=+sqrt%28+108+%29+
c+=+10.3923+
The expression for the foci of a general ellipse is
Foci: (c+h, k) (-c+h, k)
So for our ellipse the foci are (10.3923+2, 8) (-10.3923+2, 8)
(12.3923, 8) (-8.3923, 8)