Trying to clarify a conceptual gap regarding shifting and rotation of axis

Question

Find the new equation of the curve $4(x-2y+1)^2+9(2x+y+2)^2=25$ if the lines $2x+y+2=0$ and $x-2y+1=0$ are taken as the new $x$ and $y$-axis respectively.

The given lines are perpendicular, so, I thought maybe $2x+y+2$ can be replaced as $x$ and $x-2y+1$ can be replaced as $y$. So, maybe the answer would be $4x^2+9y^2=25$

But the answer given is $4x^2+9y^2=5$

Seems like they replaced $\frac{2x+y+2}{\sqrt{2^2+1^2}}$ as $x$, i.e. seems like they replaced perpendicular distances with $x$ and $y$, but why?

I am aware of shifting of origin and rotation of axis. And in this question both concepts seem to be applied.

Also, rotation of axis also involves $\sin\theta$, $\cos\theta$ etc, perhaps that's not needed here, given the staight forward answer.

Also, the fact that the given curve is ellipse has any bearing on the approach?

This question is present on infinity learn website

Answer 1

You want to make a change of variables, so that the new axes are aligned with two perpendicular lines with equations $$ ax+by+c=0,\quad a'x+b'y+c'=0. $$ If you want lengths to be the same with the new coordinates $(X,Y)$, then you must choose them as follows (or swapping $X$ with $Y$): $$ X={ax+by+c\over\sqrt{a^2+b^2}},\quad Y={a'x+b'y+c'\over\sqrt{a'^2+b'^2}}. \tag1 $$ PROOF. Note first of all that, as those two lines are perpendicular, then $aa'+bb'=0$, that is: $$ {a'\over b}=-{b'\over a}=k. $$ We can then substitute $a'=kb$ and $b'=-ka$ in equation $(1)$ obtaining: $$ X={ax+by+c\over\sqrt{a^2+b^2}},\quad Y={bx-ay+c'/k\over\sqrt{a^2+b^2}}. \tag1 $$ Take now any two points expressed by $(x_1,y_1)$ and $(x_2,y_2)$ in the old coordinates. Their squared distance, computed with the new coordinates, is: $$ d^2=(X_1-X_2)^2+(Y_1-Y_2)^2= {(ax_1+by_1-ax_2-by_2)^2\over{a^2+b^2}}+ {(bx_1-ay_1-bx_2+ay_2)^2\over{a^2+b^2}} $$ that is: $$ \begin{align} d^2 & = {a^2(x_1-x_2)^2+b^2(y_1-y_2)^2+2ab(x_1-x_2)(y_1-y_2)\over{a^2+b^2}}\\ &+{b^2(x_1-x_2)^2+a^2(y_1-y_2)^2-2ab(x_1-x_2)(y_1-y_2)\over{a^2+b^2}}\\ &=(x_1-x_2)^2+(y_1-y_2)^2. \end{align} $$ But the last expression is the squared distance computed with the old coordinates: that completes the proof.

Answer 2

First, I suspect that might be a typo of your book. Nevertheless, you can always take the new axis and scale it, for example, let

$kx'=2x+y+2$ and $ky'=x-2y+1$

now you get

$4k^2x'^2+9k^2y'^2=25$

you can take different $k$ and the curve will be scaled.

Answer 3

The expanded equation is $40x^2+20xy+25y^2+80x+20y+15=0.$ Treat this to the translate and rotate standard matrix treatment which gives a center that we translate to the origin and a rotation matrix $P=\begin{pmatrix}\cos{\theta}&-\sin{\theta}\\\sin{\theta}&\cos{\theta}\end{pmatrix}$ for one $\theta \in (-180^\circ,180^\circ].$

I suspect the ones making the exercise want you to make sure yours is a determinant $1$ one coming from such a cosine/sine rotation.

This forces the order P=matrix([1/sqrt(5),-2/sqrt(5)],[2/sqrt(5),1/sqrt(5)]) of the normalized eigenvectors as columns in the rotation matrix and 4*x'^2+9*y'^2=5. The other order would switch $x$ and $y.$

Or 4*((x+1)-2*y)^2/5+9*(2*(x+1)+y)^2/5=5 going back again through the translation from (-1,0) and rotation by 63.43... degrees.