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

1 Answer
Jim H
Sep 29, 2015

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

Explanation:

11x^2+5.5y^2-22x+11y=0

Complete the square to put this into standard form:

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

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

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

(x-1)^2/(16.5/11)+(y+1)^2/(16.5/5.5)=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+-sqrt3/2,-1) and (1, -1+-sqrt3).

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

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