# How do you find the coordinates of the center for the given ellipse 4x^2+16x+352+16y^2+160y=0?

Dec 23, 2015

$\left(- 2 , - 5\right)$

#### Explanation:

Rearrange:

$\left(4 {x}^{2} + 16 x\right) + \left(16 {y}^{2} + 160 y\right) = - 352$

Complete the square after factoring out the coefficient on the squared terms.

$\textcolor{red}{4} \left({x}^{2} + 4 x + \textcolor{red}{4}\right) + \textcolor{b l u e}{16} \left({y}^{2} + 10 y + \textcolor{b l u e}{25}\right) = - 352 + \textcolor{red}{16} + \textcolor{b l u e}{400}$

$4 {\left(x + 2\right)}^{2} + 16 {\left(y + 5\right)}^{2} = 64$

${\left(x + 2\right)}^{2} / 16 + {\left(y + 5\right)}^{2} / 4 = 1$

This is in the standard form of an ellipse:

${\left(x - h\right)}^{2} / a + {\left(y - k\right)}^{2} / b = 1$

Where the ellipse has a center at $\left(h , k\right)$.

The trick to finding the center is remembering to switch the sign of whatever's inside the parentheses:

${\left(x + 2\right)}^{2} \rightarrow x \text{-coordinate } \textcolor{\in \mathrm{di} g o}{- 2}$
${\left(y + 5\right)}^{2} \rightarrow y \text{-coordinate } \textcolor{\in \mathrm{di} g o}{- 5}$

$\left(- 2 , - 5\right)$

graph{(4x^2+16x)+(16y^2+160y)=-352 [-11.25, 8.75, -8.04, 1.96]}