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

1 Answer
Sep 29, 2015

Answer:

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]}