Posted by Honey on .
4x^2+y^248x4y+48=0 what is the distance from c to foci of this equation
The center is (6,2)
major is 10 i think
and minor is 5 but they might be wrong

Ellipses Part 2:) 
Steve,
4x^2+y^248x4y+48=0
4(x^2  12x) + (y^2  4y) = 48
4(x^2  12x + 36) + (y^2  4y + 4) = 48 + 144 + 4
4(x6)^2 + (y2)^2 = 100
(x6)^2/25 + (y2)^2/100 = 1
You are correct, although
semimajor is 10
semiminor is 5
c^2 = 100  25
c = sqrt(75) = 5*sqrt(3)