Using complex numbers to find reflections

Question

if the equation of the curve of the reflection of ellipse $ \frac{(x-4)^2}{16} + \frac{ (y-3)^2}{9} = 1$ about the line $x-y-2=0$ is $16x^2 + 9y^2 + k_1 x -36 y+k_2 =0 $ , then $ \frac{k_1 +k_2}{33}$ =?

So, I thought of this method using complex numbers to find reflection point (z'), so say I want to reflect a point $ z=(x+iy)$ about a line, then I do this sequence of transformation.

$$ z' = \overline{z} e^{ 2i \arctan(m) } $$

where 'm' is slope of line,

So, I first do a coordinate transform for the ellipse

$ x= 4 + 4 \cos \theta$

$ y = 3 + 3 \sin \theta$

And thne,

I put $ z= (4 + 4 \cos \theta) + i( 3+ 3 \sin \theta)$

$\overline{z} = (4 +4 \cos \theta) – i (3 + 3 \sin \theta)$

so, $ m= \frac{\pi}{4}$

I get,

$ z' = \overline{z} (i)$\

but this doesn't give me the right answer for some reason…

P.s: the motivation for the reflection point formula is that, first I make the line my x axis byrotating whole plane by negative of slope of line i.e : \overline{ z e^{i \arctan(m)} , then to find reflection I took reflection of this point about 'x' axis by conjugating it then I multiply it by $ e^{ i \arctan(m)} $ to find the point in original coordinate system

Where exactly am I going wrong?

If the equation of the curve on the reflection of the ellipse $\frac{(x-4)^2}{16}+\frac{(y-3)^2}{9}=1$ about the line $x-y-2=0$ is …

I saw this, but I want to do this using complex numbers
reference for my method:

1. I make my line of refleciton my axis
2. I conjugate my point
3. I rotate my point back by how much I rotated it originally

Part that I am confused with : Why do we shift the line such that intercepts line up with origin? I know we either shift x intercept to origin or y intercept to origin coz x intercept and y intercept related. But why do we start with this?

Answer 1

The diagram is on the right track, but you will also have to do a translation given the fact that $y=x-2$.

The steps as I would envision are:

1. Make the following substitutions: $$ x = \frac{z+ \overline{z}}{2}$$ $$ y = \frac{z- \overline{z}}{2i}$$

2. Apply the rotational coordinate transformation: $$ z' = ze^{-i\theta}$$ $$\theta = artcan(m)$$

3. Apply the translation coordinate transformation: $$ z'' = z' + ai $$ $$ a = \sqrt{2} $$

4. Perform the reflection: $$ z'' = \overline{z''}$$

5. 'Undo' the translation coordinate transformation: $$ z' = z'' - ai $$

6. 'Undo' the rotational coordinate transformation: $$ z = z'e^{i\theta}$$

Diagram of translation first approach (Ellipse position is very approximate):