# How do I rotate the axes of and then graph 11x^2+5.5y^2-22x+11y=0?

Sep 29, 2015

There is no $x y$ term, so it is not necessary to rotate the axes.

#### Explanation:

$11 {x}^{2} + 5.5 {y}^{2} - 22 x + 11 y = 0$

Complete the square to put this into standard form:

$\left(11 {x}^{2} - 22 x \textcolor{w h i t e}{\text{XX")+(5.5y^2+11ycolor(white)"XX}}\right) = 0$

$11 \left({x}^{2} - 2 x \textcolor{w h i t e}{\text{XXX")+5.5(y^2+2ycolor(white)"XXX}}\right) = 0$

$11 \left({x}^{2} - 2 x + 1\right) + 5.5 \left({y}^{2} + 2 y + 1\right) = 0 + 11 + 5.5$

$11 {\left(x - 1\right)}^{2} + 5.5 {\left(y + 1\right)}^{2} = 16.5$

${\left(x - 1\right)}^{2} / \left(\frac{16.5}{11}\right) + {\left(y + 1\right)}^{2} / \left(\frac{16.5}{5.5}\right) = 1$

${\left(x - 1\right)}^{2} / \left(\frac{3}{2}\right) + {\left(y + 1\right)}^{2} / 3 = 1$

The graph is an ellipse with center #(1,-1) and endpoints of axes:

$\left(1 \pm \frac{\sqrt{3}}{2} , - 1\right)$ and $\left(1 , - 1 \pm \sqrt{3}\right)$.

graph{11x^2+5.5y^2-22x+11y=0 [-4.03, 7.07, -4.313, 1.237]}