Question

$\text{Find what the following equations become when the origin is shifted to the point(1,1)}?\left(i\right)x^2+xy-3y^2-y+2=0\left(ii\right)xy-y^2-x+y=0\left(iii\right)xy-x-y+1=0\left(iv\right)x^2-y^2-2x+2y=0$

Answer 1

$\left(i\right)x^2+xy-3y^2-y+2=0\text{Substituting x = X + 1, y= Y + 1 in the equation,}\left(i\right)\text{we get}(X+1{)}^2+(X+1\left)\right(Y+1)-3(Y+1{)}^2-(Y+1)+2=0\Rightarrow X^2+1+2X+XY+X+Y+1-3(Y^2+1+2Y)-Y-1+2=0\Rightarrow X^2+XY+3X+3-3Y^2-3-6Y=0\Rightarrow X^2-3Y^2+XY+3X-6Y=0\text{(ii) We have,}xy-y^2-x+y=0...\left(i\right)\text{Substituting x = X + 1, y = Y + 1 in the equation,}\text{(i), we get}(X+1)(Y+1)-(Y+1{)}^2-(X+1)+(Y+1)=0\Rightarrow XY+X+Y+1-(Y^2+1+2Y)-X-1+Y+1=0\Rightarrow XY+2Y+1-Y^2-1-2Y=0\Rightarrow XY-Y^2=0\text{(iii) xy - x - y + 1 = 0}\text{Substituting x = X + 1, y = Y + 1 in the equation,}\text{(i), we get}(X+1\left)\right(Y+1)-(X+1)-(Y+1)+1=0\Rightarrow XY+X+Y+1-X-1-Y-1+1=0\Rightarrow XY=0\text{(iv) We have,}{x}^2-y^2-2x+2y=0\text{Substituting x = X + 1, y = Y + 1 in the equation,}\text{(i), we get}(X+1{)}^2-(Y+1{)}^2-2(X+1)+2(Y+1)=0\Rightarrow X^2+1+2X-(Y^2+1+2Y)-2X-2+2Y+2=0\Rightarrow X^2+1-Y^2-1-2Y+2Y=0\Rightarrow X^2-Y^2=0$