How do use the method of translation of axes to sketch the curve #x^2+4y^2-4x+24y+36=0#?

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Answer 1 · 2017-10-07T13:25:30

The new axes are #((X),(Y))=((x-2),(y+3))#

Let rewrite and factorise the equation of the curve

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

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

Complete the squares for #x# and #y#

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

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

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

Dividing by #4#

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

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

Let the new system of axes be #(X,Y)# such that

#X=x-2#

#Y=y+3#

Therefore,

#X^2/4+Y^2/1=1#

This is the equation of an ellipse.

graph{(x^2/4+y^2-1)((x-2)^2/4+(y+3)^2-1)=0 [-10, 10, -5, 5]}