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

1 Answer

Sep 29, 2015

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

Explanation:

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

Complete the square to put this into standard form:

$(11 x^{2} - 22 x XX) + (5.5 y^{2} + 11 y XX) = 0$ $(11x^2-22xcolor(white)"XX")+(5.5y^2+11ycolor(white)"XX")=0$

$11 (x^{2} - 2 x XXX) + 5.5 (y^{2} + 2 y XXX) = 0$ $11(x^2-2xcolor(white)"XXX")+5.5(y^2+2ycolor(white)"XXX")=0$

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

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

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

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

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

$(1 \pm \frac{\sqrt{3}}{2}, - 1)$ $(1+-sqrt3/2,-1)$ and $(1, - 1 \pm \sqrt{3})$ $(1, -1+-sqrt3)$ .

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

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