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Let $F=(0, f)$ be a point. This is the focus of the parabola $$y=\frac{x^2}{2f}+\frac{f}{2}$$ with the $x$-axis as the directrix.

I want to prove that a light ray travelling downward (parallel to the $y$-axis) is reflected from the parabola in such a way that the reflected ray travels through the focus $F$.

Let's consider the function

$$g(x,y) = y^2 - \left(x^2 + (y-f)^2 \right). $$

This is the difference of the squared distances from $(x, y)$ to the $x$-axis and to $F$ so the parabola is the level set $\{g=0\}$. Therefore the gradient of $g$ is perpendicular to the tangent of the parabola. The gradient is

$$\nabla g(x, \space y) = (0,\space 2y) - (2x, \space 2(y-f)).$$ Denote the parts $v = (0,\space 2y)$ and $w = -(2x, \space 2(y-f))$ These are vectors originating from the point $(x,y)$ ($=(x, \space \frac{x^2}{2f}+\frac{f}{2})$ since it's on the parabola). The vector $v$ points directly up and the vector $w$ towards the point $F$. These facts are clear from the formulas and also because the vectors are the gradients of the functions "squared distance to $x$-axis" and $-d((x,y), (0,f))^2$, respectively, and as such point towards the greatest increase.

gradient of g bisects the ray's reflection

So to prove the claim, let's show that $\nabla g$ bisects the angle between $v$ and $w$ (then the ray coming in the direction $-v$ is reflected along $w$, since $\nabla g$ is normal to the parabola). To prove this it suffices to show that

$$|v| = |w|$$ because $\nabla g = v+w$.

So, here we came to my actual question:

Is $|v| = |w|$ clear intuitively from the fact that these are the gradients of the distance (squared) functions and the distance grows to "the maximal growth direction" always at the same pace. (We had distance squared but could have had the square root there, it wouldn't change the directions of the greatest increase.)

I can calculate the squared length of both $v$ and $w$ to be $\frac{x^4}{f^2}+2x^2+f^2$ and hence prove $|v| = |w|$, but I'd like to understand the gradient better.

Maybe the geometric/analytic reasoning of $|v| = |w|$ could be made more formal by considering the $1$D-functions that measure the distance to the maximal direction and showing they grow at the same rate (at the rate $1$).

This is more of a proof verification type of post but I'd like to hear some thoughts of this sort of geometric-analytic proof and acquire deeper knowledge of the gradient :).

ploosu2
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1 Answers1

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The equation of the parabola can be written as $$x^2=4\left(\frac f2\right)\left(y-\frac f2\right)$$ The coordinates of point $P$ with parameter $p$ are $$\bigg(fp,\;\; \frac f2 \big(p^2+1\big)\bigg)$$ The gradients of the tangent and normal at $P$ respectively are $$p, -\frac 1p$$ The gradient of $FP$, $\tan\alpha$ is $$\frac {p^2-1}{2p}$$ The gradient of the ray $PI$ hitting $P$ and parallel to the axis of the parabola is $\infty$.

Rotate these four lines anticlockwise by $\theta$ where $\tan\theta =p$ is the gradient of the tangent at $P$.

The post-rotation gradients of the lines are:

  • tangent at P: $0$
  • PI: $\dfrac 1p$
  • normal at P: $\infty$
  • FP: $\dfrac {\tan\alpha-\tan\theta}{1+\tan\alpha\tan\theta}=\dfrac {\frac {p^2-1}{2p}-p}{1+p\left(\frac {p^2-1}{2p}\right)}=-\dfrac 1p$

The post-rotation gradient of PI and FP are equal in magnitude but opposite in sign, i.e. symmetrical about the rotated normal.

Hence, in the unrotated case, both PI and FP are also symmetrical about the normal.

Hypergeometricx
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