Question 66246
I'm afraid the other solution that was posted for this problem is incorrect.  Anytime you have an equation of a conic section with {{{x^2}}} and {{{y^2}}} with both having positive but unequal coefficients, it will be an ELLIPSE, NOT a PARABOLA. 


If both coefficients are the same sign with equal coefficients, it will be a CIRCLE.  


If coefficients are of opposite sign, then it will be a HYPERBOLA.  


A PARABOLA results when there is an {{{x^2}}} but no {{{y^2}}} or a {{{y^2}}} but no {{{x^2}}}.  


This is the way it should be solved by completing the square:
{{{4x^2-16x+9y^2+18y=0}}}


You must first factor out the coefficients of x^2 and y^2 like this:
{{{4(x^2-4x+____)+9(y^2 +2y + _____)=0}}}

{{{4(x^2-4x+4)+9(y^2 +2y + 1)=0+16+9}}}
{{{4(x^2-4x+4)+9(y^2 +2y + 1)=25}}}


This is an ellipse.  The standard form for an ellipse is in a form = 1, so divide both sides of the equation by 25 to set it equal to 1.

{{{(4(x-2)^2)/25 +(9(y+ 1)^2)/25=25/25}}}
{{{(4(x-2)^2)/25 +(9(y+ 1)^2)/25=1}}}

Finally invert the coefficients of 4 and 9 in order to write this:
{{{((x-2)^2)/(25/4) +((y+ 1)^2)/(25/9)=1}}}


That would be standard form for an ellipse!  The center is at (2,-1), with the "radius" extending {{{5/2}}} units in the x direction, and {{{5/3}}} units in the y direction. 

{{{graph (300,300,-5,5,-5,5, -1+(sqrt(25-4(x-2)^2))/3, -1-(sqrt(25-4(x-2)^2))/3) }}}


R^2 at SCC