Question

If the origin is shifted to the point (1, -2) without rotation of axes what do the following equations become?

i. $2x^2 + y^2 -4x + 4y = 0$ and ii. $y^2 - 4x + 4y + 8 = 0$.

Answer 1

i. $2x'^2 + y'^2 - 6 = 0$ and ii. $y'^2 - 4x' = 0$

Answer 2

The origin is shifted from (0,0) to (1, -2). The transformation equations are $x = x' + 1$ and $y = y' - 2$, where $(x, y)$ are the original coordinates and $(x', y')$ are the new coordinates.

i. For $2x^2 + y^2 -4x + 4y = 0$: $2(x'+1)^2 + (y'-2)^2 - 4(x'+1) + 4(y'-2) = 0$ $2(x'^2 + 2x' + 1) + (y'^2 - 4y' + 4) - 4x' - 4 + 4y' - 8 = 0$ $2x'^2 + 4x' + 2 + y'^2 - 4y' + 4 - 4x' - 4 + 4y' - 8 = 0$ $2x'^2 + y'^2 - 6 = 0$

ii. For $y^2 - 4x + 4y + 8 = 0$: $(y'-2)^2 - 4(x'+1) + 4(y'-2) + 8 = 0$ $(y'^2 - 4y' + 4) - 4x' - 4 + 4y' - 8 + 8 = 0$ $y'^2 - 4y' + 4 - 4x' - 4 + 4y' = 0$ $y'^2 - 4x' = 0$

The transformed equations are: i. $2x'^2 + y'^2 - 6 = 0$ ii. $y'^2 - 4x' = 0$