# Reflection of a curve/point using graph transformations.

The original question is to reflect the curve $$y^2=4ax$$ about the line $$y+x=a$$.

The general method to solve such a question is to consider the parametric coordinates of the given curve (in this case $$(at^2,2at)$$) and reflect this general point about the given line and then eliminate the parameter from these reflected coordinates to get the curve.

But in this case I used graph transformations. Notice that the line already has the slope of $$-1$$

(this struck me because reflection of any point about $$y=\pm x$$ are sort of standard results, involving just the simple transformation {$$x_1 = y$$ and $$y_1= x$$} or {$$x_1= -y$$ and $$y_1= -x$$} for $$y=x$$ and $$y=-x$$ respectively.)

Here is the method I used:

First allow $$y= y_1+a$$. This makes the line equation into $$y=-x$$ and the curve becomes $$(y+a)^2 = 4ax$$. Now in this frame we reflect the curve about the line, which just means $$y= -x_1$$ and $$x=-y_1$$. This means the reflected curve is $$(a-x)^2 = -4ay$$. Now we can shift all the thing back which means $$y= y_1-a$$ which makes the final reflected curve $$(a-x)^2 =4a(a-y)$$. (which is the correct answer by the way).

So I wondered if I could use these types of transformations in reflection about a line with any slope. I stated out with points first because its easier to verify using points rather than curves.

Consider $$P(4,7)$$. Reflect it about $$x+y=7$$. I used the same logic with similar transformations and got the correct reflected point.

But when I tried to reflect it about $$y=2x$$, look what happens.

First $$x= x_1/2$$ which makes the line $$y=x$$ and $$P'(8,7)$$. Now reflection about $$y=x$$ gives us $$P''(7,8)$$ and then transforming it back, using $$x=2x_1$$, we get $$P_{ref}(7/2 , 8)$$ which is not the right reflected coordinate. (actual reflected point is $$(16/5 , 37/5)$$).

So my questions are:

1. Why is the transformation method not working if the slope of the line is not $$\pm 1$$?

2. Is there any way to make this method work ?

• This is an interesting question and I will look into it. However, your use of $y\to y+a$ is rather confusing and it would probably be better to just introduce a new variable $y'=y+a$. Jul 1 '20 at 15:56
• Thank you for the suggestion, I have now edited the content. Jul 1 '20 at 17:28
• @K.defaoite that just adds to the confusion as ‘ signifies a derivative. Jul 1 '20 at 17:52
• @Radial Arm Saw I edited the content once again, replacing $x'$ by $x_1$. Jul 1 '20 at 19:16
• Much of the time in mathematics, the $'$ does not signify a derivative. In fact you'll find as you get further in math, the prime notation for derivatives is usually to be avoided if possible. Jul 1 '20 at 19:41

Sketch the line $$L: y=2x$$, a point, $$P$$, and its reflection across that line, $$R$$. The segment between the two points intersects the line at right angles in some point $$Q$$.
Apply the stretching transformation $$x_{str}=2x$$ to all of these objects. You'll see that $$Q_{str}$$ is indeed the midpoint between $$P_{str}$$ and $$R_{str}$$, but that the segment joining those points does not meet the line $$L_{str}$$ at right angles.
If you try to reflect $$P_{str}$$ across $$L_{str}$$, you will not reach the correct point $$R_{str}$$.
To solve the problem, transform the points in a way that does preserve angles, like by rotating the space around the origin to map the line of reflection onto $$y=x$$.