4x^2+y^2-48x-4y+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
4x^2+y^2-48x-4y+48=0
4(x^2 - 12x) + (y^2 - 4y) = -48
4(x^2 - 12x + 36) + (y^2 - 4y + 4) = -48 + 144 + 4
4(x-6)^2 + (y-2)^2 = 100
(x-6)^2/25 + (y-2)^2/100 = 1
You are correct, although
semi-major is 10
semi-minor is 5
c^2 = 100 - 25
c = sqrt(75) = 5*sqrt(3)