# How do you change x^2 + 4y^2 - 4x + 8y - 60 = 0 into standard form?

Mar 20, 2016

$\frac{{\left(x - 2\right)}^{2}}{{\left(2 \sqrt{17}\right)}^{2}} + \frac{{\left(y + 1\right)}^{2}}{{\left(\frac{\sqrt{17}}{2}\right)}^{2}} = 1$

#### Explanation:

The given equation is already in standard polynomial form, so I assume that what is required is to express this in standard elliptical form.
$\textcolor{w h i t e}{\text{XXX}} \frac{{\left(x - h\right)}^{2}}{{a}^{2}} + \frac{{\left(y - k\right)}^{2}}{{b}^{2}} = 1$

Given:
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} + 4 {y}^{2} - 4 x + 8 y - 60 = 0$

Re-arranging the terms to group the $x$ terms, the $y$ terms, and shift the constant to the right side:
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} - 4 x + 4 {y}^{2} + 8 y = 60$

Extract the common factor from the $y$ terms and complete the squares
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} - 4 x + 4 + 4 \left({y}^{2} + 2 y + 1\right) = 60 + 4 + 4$

Re-writing as squared binomials and simplifying
$\textcolor{w h i t e}{\text{XXX}} {\left(x - 2\right)}^{2} + 4 {\left(y + 1\right)}^{2} = 68$

Divide through by 68and take roots of denominators
$\textcolor{w h i t e}{\text{XXX}} \frac{{\left(x - 2\right)}^{2}}{{\left(2 \sqrt{17}\right)}^{2}} + \frac{{\left(y + 1\right)}^{2}}{{\left(\frac{\sqrt{17}}{2}\right)}^{2}} = 1$