Precalculus
posted by Emily .
"Sketch the ellipse. Find the coordinates of its vertices and foci.
34. x^2 + 25y^2  6x  100y + 84 = 0
(Hint: Complete the squares in x and y. Begin by writing the equation in this form: (x^2  6x + ___) + 25(y^2  4y + ___) = 84.)"
==> I used the hint that came with the problem, but as I tried to solve the equation into the general form for an ellipse, I was getting negative denominators, which would make a and b contain imaginary numbers. Does anyone know how to solve this? Any help is greatly appreciated! :)

x^2 + 25y^2  6x  100y + 84 = 0
x^  6x + 9 + 25(y^2  4y + 4) = 84 + 9 + 100
notice in the y expression, I added 4 to the inside, but that was multiplied by 25, so I had to add 100 on the right side.
(x3)^2 + 25(y2)^2 = 25
(x3)^2/25 + (y2)^2 = 1
so a^2 = 25, and b^2 = 1
take it from there 
Ohhhhh!!!!! This totally makes sense. I forgot to multiply the 4 by 25 before bringing it over to the other side. Thanks so much!! :D