find the focus of x^2+4y^2+2x+24y++28=0. Thank you
x^2+4y^2+2x+24y+28=0
x^2+2x + 4y^2+24y + 28 = 0
x^2+2x+1 + 4(y^2+6y+9) + 28-1-4*9 = 0
(x+1)^2 + 4(y+3)^2 = 9
(x+1)^2/9 + (y+3)^2/(9/4) = 1
Now we see that a=3, b=3/2, so c = 3√3/2
the foci are at (-1±3√3/2,-3)
see
http://www.wolframalpha.com/input/?i=ellipse+x%5E2%2B4y%5E2%2B2x%2B24y%2B28%3D0