Bobosharif S. answered • 06/07/18

Mathematics/Statistics Tutor

3x

^{2}+y^{2}-6x+4y+4=03x

^{2}-6x+y^{2}+4y+4=03(x

^{2}-2x+1-1)+(y^{2}+4y+4)=03(x-1)

^{2}-3+(y+2)^{2}=03(x-1)

^{2}+(y+2)^{2}=3 (divide both sides by 3)(x-1)

^{2}/1+(y+2)^{2}/3=1This is a canonical form of ellipse equation.