How do you find the standard form given #9x^2+4y^2-36x+24y+36=0#?

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Answer 1 · 2017-02-15T06:52:46

The answer is #(x-2)^2/4+(y+3)^2/9=1#

Let's do some rearrangement by completing the squares

#9x^2+4y^2-36x+24y+36=0#

#9x^2-36x+4y^2+24y=-36#

#9(x^2-4x)+4(y^2+6y)=-36#

#9(x^2-4x+4)+4(y^2+6y+9)=-36+36+36#

#9(x-2)^2+4(y+3)^2=36#

#(x-2)^2/4+(y+3)^2/9=1#

This is the equation of an ellipse.

graph{9x^2-36x+4y^2+24y+36=0 [-11.41, 11.09, -11.1, 0.15]}