Problem With Transformation of Coordinates

Question

In S.L. Loney's book on Coordinate Geometry, he introduces the reader a way to switch from one origin of coordinates to another. The method he shows is: to change the origin to the point $(h,k)$ you must replace the quanitities of $x$ and $y$ with $x + h$ and $y + k$. However, using an online graphic calculator, the examples he shows don't seem correct to me. According to his method, say I want to change the origin of coordinates to the point $(2,2)$; so if I have the curve $y^2 = 4x$ referred to in previous coordinates, the new curve should be $(y + 2)^2 = 4(x + 2)$. However, when I check the online graphic calculator, the method has changed the origin to $(-2,-2)$, not $(2,2)$. Could someone explain what I am misinterpretting? Thank you in advance.

Answer 1

Let’s look at Ex. 1 on p. 111 of The Elements of Coordinate Geometry by Sidney Luxton Loney (1895), which is the example you bring up in your comments.

Transform to parallel axes through the point $(-2,3)$ the equation $$2x^2 \; + \; 4xy \; + \; 5y^2 \; - \; 4x \; - \; 22y \; + \; 7 \; = \;0.$$

The original variables are $x,y$ and the new variables are $x',y'.$ Each point has both an $x,y$ coordinate description and an $x',y'$ coordinate description.

We want the new origin to be the point whose $xy$ description is given by $(x,y) = (-2,3).$ That is, we want $(x,y) = (-2,3)$ and $(x',y')=(0,0)$ to be descriptions of the same point. Since we want the coordinate changes to be “as simple as possible” to achieve this (here I'm using hand-waving wording), we do this as follows:

$$ \text{add} \; 2 \; \text{to} \;\; x=-2 \;\; \text{to get} \; x’=0$$ $$ \text{subtract} \; 3 \; \text{from} \;\; y=3 \;\; \text{to get} \; y’=0$$

That is, to get $x'$ we add $2$ to $x,$ and to get $y'$ we subtract $3$ from $y.$

$$ x' = x + 2 \;\;\; \text{or} \;\;\; x = x' - 2 $$ $$ y' = y - 3 \;\;\; \text{or} \;\;\; y = y' + 3 $$

Thus, given that $2x^2+4xy+5y^2-4x-22y+7=0$ is the $xy$-equation of the locus, then the $x'y'$-equation of the same locus is obtained by replacing $x$ with $x’-2$ and replacing $y$ with $y’+3.$ This gives

$$2(x'-2)^2 \; + \; 4(x'-2)(y'+3) \; + \; 5(y'+3)^2 \; - \; 4(x'-2) \; - \; 22(y'+3) \; + 7 \;\; = \;\; 0 $$

$$2(x'^2-4x'+4) + 4(x'y'+3x'-2y'-6) + 5(y'^2+6y'+9) - 4x' + 8 - 22y' - 66 + 7 \; = \; 0 $$

$$2x'^2-8x'+ 8 + 4x'y'+12x'-8y'-24 + 5y'^2+30y'+45 - 4x' + 8 - 22y' - 66 + 7 \; = \; 0 $$

$$2x'^2 + 4x'y' + 25y'^2 + (-8+12-4)x' + (-8+30-22)y' + (8-24+45+8-66+7) \;\; = \;\; 0 $$

$$2x'^2 \; + \; 4x'y' \; + \; 25y'^2 \; - \; 22 \;\; = \;\; 0 $$