Question

put this parabola in standard form ; 5x^2+12y^2+20x+96y+152=0

Answer 1

This is not a parabola. You might be confusing this term with "conic section."
\[5x^2+12y^2+20x+96y+152=0\]
Complete the square:
\[\begin{align*}5\left(x^2+4x\right)+12\left(y^2+8y\right)+152&=0\\
5\left(x^2+4x+4-4\right)+12\left(y^2+8y+16-16\right)+152&=0\\
5\left((x+2)^2-4\right)+12\left((y+4)^2-16\right)+152&=0\\
5(x+2)^2-20+12(y+4)^2-192+152&=0\\
5(x+2)^2+12(y+4)^2-60&=0\\
&\vdots
\end{align*}\]
Feel to solve for \(y\) or \(x\), or just leave it in the standard form of an ellipse (the conic section that this equation represents): \(\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1\), where \((h,k)\) is the center of the ellipse, \(a\) is the length of the horizontal axis, and \(b\) is the length of the vertical axis.

Here's an image: http://www.wolframalpha.com/input/?i=5x%5E2%2B12y%5E2%2B20x%2B96y%2B152%3D0