# How do you find the standard form of x^2 + 4y^2 - 8x + 16y + 16 =0 and what kind of a conic is it?

Dec 12, 2015

An ellipse with center $\left(4 , - 2\right)$, x-axis radius: $4$, and y-axis radius: $2$

#### Explanation:

Given:
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + 4 {y}^{2} - 8 x + 16 y + 16 = 0$

Group $x$ and $y$ terms (and constant) separately as
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l u e}{{x}^{2} - 8 x} + \textcolor{red}{4 \left({y}^{2} + 4 y\right)} + \textcolor{g r e e n}{16} = 0$

Complete the squares for both the $x$ and the $y$ sub-expressions:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l u e}{{x}^{2} - 8 x + 16} + \textcolor{red}{4 \left({y}^{2} + 4 y + 4\right)} + \textcolor{g r e e n}{16} - \textcolor{b l u e}{16} - \textcolor{red}{4 \left(4\right)} = 0$

Reduce $x$ and $y$ sub-expressions to squared binomials and move constant to the right side:
$\textcolor{w h i t e}{\text{XXX}} {\left(x - 4\right)}^{2} + 4 {\left(y + 2\right)}^{2} = 16$
or
$\textcolor{w h i t e}{\text{XXX}} {\left(x - 4\right)}^{2} + 4 {\left(y + 2\right)}^{2} = {4}^{2}$

Divide both sides by $\left({4}^{2}\right)$
and replace $\left(y + 2\right)$ with $\left(y - \left(- 2\right)\right)$
$\textcolor{w h i t e}{\text{XXX}} \frac{{\left(x - 4\right)}^{2}}{{4}^{2}} + {\left(y - \left(- 2\right)\right)}^{2} / \left({2}^{2}\right) = 1$

Note that the general form of an ellipse is
$\textcolor{w h i t e}{\text{XXX}} {\left(x - h\right)}^{2} / \left({a}^{2}\right) + \frac{{\left(y - k\right)}^{2}}{{b}^{2}} = 1$
with
$\textcolor{w h i t e}{\text{XXX}}$center at $\left(h , k\right)$,
$\textcolor{w h i t e}{\text{XXX}}$x-axis radius: $\left(a\right)$, and
$\textcolor{w h i t e}{\text{XXX}}$y-axis radius: $\left(b\right)$
graph{x^2+4y^2-8x+16y+16=0 [-2.195, 10.29, -4.895, 1.345]}