I've got the circles and parabolas but I am having trouble with these... any guidance would be greatly appreciated.

Ellipse: x^2 + 4y^2 + 6x + 16y + 21 = 0
(x^2 + 6x +9) + 4y^2+16y) = -21 +9
(x+3)^2 + 4(y^2 + 4y+4) = -12 + 16
(x+3)^2 + 4(y+2)^2=4
divide both sides by 4
(x+3)^2 / 2^2 + (y+2)^2 / 1^2= 1
Center (-3,-2)
What next? Is the above correct?

Hyperbola: 4y^2-9x^2+8y-54x-81=0
(4y^2+8y ) - (9x^2-54x ) = 81
4(y^2+2y+1) - 9(x^2-6x+9) = 81+4-81
4(y+1)^2 - 9(x-3)^2=4
divide both sides by 4
(y+1)^2 / 1^2 - 9(x-3)^2 / 2^2 = 1
here I wasn't sure what to do with the 9

Your ellipse looks perfect. The next step might be to graph it.

Your hyperbola has a mistake in it. Going from line 1 to line 2 you forgot to change -54x to +54x since you factored out the negative sign outside the brackets.

Yes but how do I get rid of the 9 on the second side? I ended up with 9(x+3)^2 / 4 I'm not sure how to handle that?

You don't need to get rid of the 9. The standard form you want to put this in is like this:


\frac {(x-a)^2} {A^2} - \frac {(y-b)^2} {B^2} = \pm 1


where (a,b,) is the center of the hyperbola, A is the semi-major axis and B is the semi-minor axis. If you combine the 9 from the denominator and the 4 in the numerator into a single denominator you have 4/9, or (2/3)^2. So 2/3 is the semi-major axis of the hyperbola, and 1 is the semi-minor axis.