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 ?

  • 2
    $\begingroup$ 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 $y'=y+a$. $\endgroup$
    – K.defaoite
    Jul 1 '20 at 15:56
  • 1
    $\begingroup$ Thank you for the suggestion, I have now edited the content. $\endgroup$
    – Aditya
    Jul 1 '20 at 17:28
  • 2
    $\begingroup$ @K.defaoite that just adds to the confusion as ‘ signifies a derivative. $\endgroup$ Jul 1 '20 at 17:52
  • 1
    $\begingroup$ @Radial Arm Saw I edited the content once again, replacing $x'$ by $x_1$. $\endgroup$
    – Aditya
    Jul 1 '20 at 19:16
  • $\begingroup$ 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. $\endgroup$
    – K.defaoite
    Jul 1 '20 at 19:41

I think that the problem here is that angles between lines are not preserved in the stretching transformation that you used.

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 $y=x$.

Of course, if you're going to do that, then it's probably easier just to rotate the line of reflection onto the x-axis, flip the sign of the point's y-value, and rotate back.


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