# Proof: Parabola reflects light to the focus

1. May 13, 2014

1. The problem statement, all variables and given/known data
The title.

2. Relevant equations
$y=mx+c$
$$\frac{dy}{dx}(kx^2)=2kx$$

3. The attempt at a solution
https://www.physicsforums.com/attachment.php?attachmentid=69745&stc=1&d=1399976415
I have proved that the angle of reflection will be a° since the angle of incidence=a° The picture will speak for itself.
The red line is the normal,perpendicular to the green tangent.

But how do I prove that it reflects light to the focus, which is (0,0.25)?

Last edited: May 13, 2014
2. May 13, 2014

### maajdl

The parabola DOES NOT REFLECT light to the focus.

For example, any ray issued from the focus never goes back to the focus.

3. May 13, 2014

Oh. I forgot to mention that the light rays are coming vertically down

4. May 13, 2014

### maajdl

You need to be more precise.
You might solve your problem if you care to formulate it clearly.

5. May 13, 2014

### HallsofIvy

I have no idea what "y= mx+ c" has to do with this problem.

Your parabola is $y= kx^2$, is that right?

A vertical line has x= a for constant a:

1) Where does the line x= a intersect the parabola?

2) What is the equation of the tangent line at that point?

3) What angle does the line x= a make with the tangent line?

4) What is the equation of the other line making that same angle with the tangent line?

5) Where does that line cross the y-axis?

(Since the slope of a line is the tangent of the angle it makes with the x-axis, you may want to use the fact that $tan(\theta+ \phi)= \frac{tan(\theta)+ tan(\phi)}{1+ tan(\theta)tan(\phi)}$)

(When I first did this kind of problem, I found it easier to put the parabola's axis along the x-axis, so that $y= \sqrt{x}$ and then use a horizontal line y= c.)

6. May 13, 2014

It has to do with this:
$(f(a),a)$ which is $(ka^2,a)$
$y=mx+c$
$y=2ka(x)+c$
$ka^2=2ka(x)+c$
$y=2ka(x)+ka^2-2ka(x)$
$y=ka^2$
What? Am I missing something here?

7. May 13, 2014

I saw the picture before. I just deleted the cache and saw that the picture was broken.Well, I don't need the picture anymore!

8. May 13, 2014

### HallsofIvy

NO. The vertical line x= a crosses the parabola $y= kx^2$ at y= $ka^2$. That is the point $(a, ka^2)$. But having x and y reversed did not affect the rest.

Okay, it is correct that the slope at x= a is 2ka.

No, $y= ka^2$ only at x= a. You should have $ka^2= 2ka(a)+ c$
so that $c= -ka^2$

You missed the fact that the tangent line touches the parabola only at x= a.
$y= 2kax- ka^2$
The important thing is that the slope (so the tangent of the angle the tangent line make with the horizontal) is 2ka. So the vertical line, x= a, makes angle $\theta$ with the tangent line so that $cot(\theta)= 2ka$

9. May 13, 2014

Oh. I reversed the x and y by mistake.
How did $ka^2= 2ka(a)+ c$ suddenly become $y= 2kax- ka^2$?
Shouldn't it be $y= 2ka^2- ka^2$?

And what is cot? Why cant we use tan there since $\tan(\theta)=\frac{\Delta y}{\Delta x}$?

10. May 13, 2014

### HallsofIvy

The equation $ka^2= 2ka(a)+ c$ is true only at x= a. I used that to find the "c" in
y= mx+ c= 2kax+ c: From $ka^2= 2ka^2+ c$, $c= -ka^2$.

11. May 14, 2014

Ok. Let's do it again!
$(a,ka^2)$
$y=2kax-kx^2$
$|180-[90+\tan^{-1}(2ka)]|$
That's difficult. Any hints?

12. May 14, 2014

### maajdl

First let's state the problem decently:

- equation of the parabola: y = k x²
- equation of an incident light ray parallel to the axis of the parabola: x = a
- question: show that the incident light ray is reflected in the direction of the focus

Let's now bite the cake.

direction of the incident light ray: d1 = (0, 1)
position of the reflection point: p = (a, ka²)
direction of the tangent to the mirror at the reflection point: t = (1, 2ka)
direction of the normal to the mirror at the reflection point: n = (-2ka,1)
direction of the reflected ray: d2 = (m, n)
how to determine (m, n):
first: d2 has the same projection along t than d1
second: d2 has the opposite projection along n compared to d1

Justify all the steps above, explain the reasoning.
Then solve for m and n.
Then write the equation of the reflected ray.
Then calculate where the reflected ray crosses the y axis.
The result should be simple: the focus.

13. May 14, 2014

I don't really understand what you meant here.

14. May 14, 2014

### HallsofIvy

This is difficult to understand. It would be better to distinguish between vectors (your "direction") and points.

The vector $\vec{j}$ or <0, 1>

and this is a point.

the derivative of $y= kx^2$ is $y'= 2kx$, and a vector in that direction, at x= a is $$\vec{i}+ 2ka\vec{j}$$ or <1, 2ka>

The normal has 0 dot product with the tangent vector and <1, 2ka>.<-2ka, 1>= 0.

The vector is <m, n> for some numbers m and n.

15. May 14, 2014

Oh my god. Vectors made it more difficult. I'm in Gr.10 so this problem is not even in my syllabus but I know differential calculus so I thought I can solve this.

16. Aug 2, 2014

### PeterBaum

Two proofs of this reflective property

You can find the essence of 2 different proofs of this property at http://en.wikipedia.org/wiki/Parabola#Proof_of_the_reflective_property

Here is a more complete proof using the diagram for the first proof shown at that site:

1. Draw FC and label its intersection with AD as point B
2. The point A on the parabola goes through the origin and is equidistant to the focus and directrix. Since the directrix is parallel to the X axis and through point C, AF=DC (this last expression is shorthand for "the length of line segment AF is equal to the length of line segment DC")
3. Angle FBA = angle DBC (opposite angles are equal)
4. Right Triangles FAB and BDC are congruent (essentially angle-side-angle)
5. AB=BD (corresponding sides of congruent triangles)
6. FB=BC (corresponding sides of congruent triangles)
7. EF = EC (all points on the parabola must be equidistant from the focus and the directrix)
8. Triangle FEB is congruent to triangle EBC (side-side-side)
9. Let the coordinates of E be (X,kX2). Then in triangle EBD, BD = X/2 (see step 5) and DE = kX2. The slope of BE is then the length of segment DE divided by the length of segment BD, which is equal to 2kX. We previously showed that the slope of the tangent through point E was also 2kX (using calculus). Therefore, line BE is the tangent of the parabola through E.
10. The angle of incidence = angle BED because they are equal opposite angles
11. Because triangle FEB is congruent to triangle BED, angle BED = angle FEB
12. The angle of incidence = angle FEB by the two previous steps. Thus the angle of reflection (originally created by drawing a line from E to the focus F is equal to the angle of incidence. Thus the lines drawn correspond to the physics rules about rays of light reflecting from a surface.

Last edited: Aug 2, 2014