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Is there a well-known formula to find $p, q, r$ in the equation $px+qy+r=0$ of the line which is the reflection of a line $ax+by+c=0$ on an axis $dx+ey+f=0$ ?

I vaguely see a way by computing angles and using trigonometry (probably tan and arctan involved), but before doing this complex method, I wanted to know if there is a ready-to-use formula.

My goal is to observe mathematically that every vertical "ray" hitting a parabola mirror will converge on a single ("focal") point:

enter image description here

See also Pure geometry proof of parabolic mirror property.

Basj
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2 Answers2

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I found the answer of the first part of my question in The equation of a line reflected about another line.

About the second part, here is a graph simulation using the formula mentioned above, drawn with Python:

enter image description here

We can see the convergence of the reflections of all vertical lines into a single ("focal") point.

import numpy as np
import matplotlib.pyplot as plt

f = lambda x: 0.2 * x ** 2
fpr = lambda x: 0.4 * x     # derivative


X = np.linspace(-3, 3, 100)
Y = f(X)


plt.plot(X, Y, linewidth=3)


for x0 in [-1, 1, 2]:


# tangent   D x + E y = F
tang = fpr(x0) * (X - x0) + f(x0)    # y = f'(x0) (x-x0) + f(x0) = f'(x0) x + f(x0) - x0 f'(x0)
l = plt.plot(X, tang, alpha=0.5, linestyle="--")
c = l[0].get_color()
D, E, F = -fpr(x0), 1, f(x0) - x0 * fpr(x0)

# vertical line  A x + B y = C
plt.axvline(x=x0, color=c)
A, B, C = 1, 0, x0

# reflection https://math.stackexchange.com/questions/67024/the-equation-of-a-line-reflected-about-another-line
k = (C * E - B * F) / (A * E - B * D)
h = (C * D - A * F) / (B * D - A * E)
a = (A * E ** 2 - A * D ** 2 - 2 * B * D * E) / (B * E ** 2 - B * D ** 2 + 2 * A * D * E)
refl = a * (X - k) + h
plt.plot(X, refl, color=c)




plt.ylim([-2, 10])
plt.axis('scaled')
plt.show()
Basj
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    I wrote an answer which proves the formula. According to the image of Wolfram Alpha, the formula looks correct where $A=4,B=1,C=8,D=1,E=-1,F=-2$ and then the formula gives $y=-\frac 14x+\frac 72$. – mathlove Jan 28 '24 at 14:44
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    Thanks @mathlove for the proof! I now understand why my plot was incorrect first: the scale was not identical for both axis, producing a weird effect on the plot. Fixed now :) – Basj Jan 28 '24 at 18:36
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The general formula for reflection $P'(x_2,y_2)$ of a point $P(x_1,y_1)$ along a line $ax+by+c=0\ $ is

$$ {x_2-x_1\over a}={y_2-y_1\over b}=-2\left({ax_1+by_1+c\over a^2+b^2}\right) $$

A general point on the line $px+qy+r=0$ is $\left({t\over p},{r-t\over q}\right)$

Which means the reflection would be $$\left({t\over p}-2a\left({\frac{at}{p}+b(\frac{r-t}{q})+c\over a^2+b^2}\right),{r-t\over q}-2b\left({\frac{at}{p}+b(\frac{r-t}{q})+c\over a^2+b^2}\right)\right)$$

The locus does give a line but it is a massive pain to solve

Its easier to see the reflection property by using the angle between lines

Assuming the parabola to be $y^2=4ax$ and a light ray $y=b$ the point of incidence being $\left({b^2\over4a},b\right)$

The normal at that point is $y=mx-2am-am^3$ where $m=-{y\over2a}=-{b\over2a}$

The slope of the line joining the point of incidence to the focus is $m={4ab\over b^2-4a^2}$

Using routine co-ord geo you can see that the angle bw the normal is and ray is half the angle bw the incident ray and reflected ray

RandomGuy
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