Question

Find what the following equations become when the origin is shifted to the point (1, 1).
(i) x² + xy − 3y² − y + 2 = 0
(ii) xy − y² − x + y = 0
(iii) xy − x − y + 1 = 0
(iv) x² − y² − 2x + 2y = 0

Answer 1

(i) The given equation is x² + xy − 3y² − y + 2 = 0.

Substituting $x=X+1,y=Y+1$ in the given equation, we get:

$$\begin{gathered} {\left(X+1\right)}^2+\left(X+1\right)\left(Y+1\right)-3{\left(Y+1\right)}^2-\left(Y+1\right)+2=0 \\ \Rightarrow X^2+1+2X+XY+X+Y+1-3Y^2-3-6Y-Y-1+2=0 \\ \Rightarrow X^2+XY-3Y^2+3X-6Y=0 \end{gathered}$$

Hence, the transformed equation is $x^2+xy-3y^2+3x-6y=0$.

(ii) The given equation is xy − y² − x + y = 0.

Substituting $x=X+1,y=Y+1$ in the given equation, we get:

$$\begin{gathered} \left(X+1\right)\left(Y+1\right)-{\left(Y+1\right)}^2-\left(X+1\right)+\left(Y+1\right)=0 \\ \Rightarrow XY+X+Y+1-Y^2-2Y-1-X-1+Y+1=0 \\ \Rightarrow XY-Y^2=0 \end{gathered}$$

Hence, the transformed equation is $xy-y^2=0$.

(iii) The given equation is xy − x − y + 1 = 0.

Substituting $x=X+1,y=Y+1$ in the given equation, we get:

$$\begin{gathered} \left(X+1\right)\left(Y+1\right)-\left(X+1\right)-\left(Y+1\right)+1=0 \\ \Rightarrow XY+X+Y+1-X-1-Y-1+1=0 \\ \Rightarrow XY=0 \end{gathered}$$

Hence, the transformed equation is $xy=0$.

(iv) The given equation is x² − y² − 2x + 2y = 0.

Substituting $x=X+1,y=Y+1$ in the given equation, we get:

$$\begin{gathered} {\left(X+1\right)}^2-{\left(Y+1\right)}^2-2\left(X+1\right)+2\left(Y+1\right)=0 \\ \Rightarrow X^2+2X+1-Y^2-2Y-1-2X-2+2Y+2=0 \\ \Rightarrow X^2-Y^2=0 \end{gathered}$$

Hence, the transformed equation is $x^2-y^2=0$.