Visual walkthrough — Reflective property of parabola (application in telescopes, antennas)

We will use only these building blocks, each introduced when needed:

• a curve (the parabola),
• a special point called the focus,
• a straight line called the directrix,
• a tangent (the flat mirror at the touch-point),
• and the law of reflection (bounce angles are equal).

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Step 1 — Draw the curve and name its parts

WHAT. A parabola is the set of points that are equally far from a fixed point and a fixed line. The fixed point is the focus $F$. The fixed line is the directrix. The mirror-line of the whole shape is the axis (here, the horizontal $x$-axis). The single point where the curve crosses its own axis — its turning point — is the vertex; for $y^2=4ax$ the vertex sits at the origin $(0,0)$.

WHY. Every later step leans on this "equal distance" rule. If we don't fix it in a picture first, symbols like $F$ and $M$ would appear out of nowhere — and this page refuses to do that.

PICTURE. Look at the figure. The green curve is the parabola $y^2=4ax$. The coral dot is the focus $F=(a,0)$. The dashed lavender vertical line on the left is the directrix $x=-a$. The slate dot at the origin is the vertex. The number $a$ is just the distance from the vertex out to $F$ — and the same distance back to the directrix.

Here $y^2=4ax$ is the standard parabola (see Parabola - standard equation y^2=4ax), and $F$, the directrix come from Focus and Directrix of a Conic.

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Step 2 — Send in the incoming ray and mark the hit-point $P$

WHAT. Fire one ray straight in from the right, travelling horizontally — parallel to the axis. It strikes the curve at a point we name $P=(x_1,y_1)$.

WHY. This horizontal ray is exactly the light from a distant star or a satellite: so far away that all its rays are parallel. We want to know where it goes after it hits the mirror.

PICTURE. The coral horizontal arrow is the incoming ray. It touches the curve at the coral dot $P$. Notice the ray runs toward the directrix — extend it and it would hit the directrix at the point $M$ from Step 1. Hold that thought: the incoming ray and the segment $PM$ point along the same horizontal line.

Here:

• $x_1$ ::: the horizontal position of the hit-point $P$ (always $\ge 0$ on this parabola).
• $y_1$ ::: the height of $P$ (we take $y_1>0$; the $y_1<0$ mirror case is Step 7).

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Step 3 — Build the tangent (the tiny flat mirror at $P$)

WHAT. A curved mirror, up close at one point, behaves like a flat mirror lined up with the tangent — the straight line that just grazes the curve at $P$. We need its slope (its steepness).

WHY. Reflection is defined against a flat surface. So to bounce the ray we must first find the flat line hugging the curve at $P$. The tool that measures "steepness of a curve at a point" is the derivative $\frac{dy}{dx}$ — that is why a derivative shows up here and nothing else would do.

HOW (term by term). Start from the curve and differentiate both sides with respect to $x$: $\underset{\text{chain rule on }{y}^2}{\underbrace{2y}}\cdot \underset{\text{slope we want}}{\underbrace{\frac{dy}{dx}}}=\underset{\text{derivative of }4ax}{\underbrace{4a}}.$ Solve for the slope: $\frac{dy}{dx}=\frac{4a}{2y}=\frac{2a}{y}.$ So at our point the tangent slope is $m_t=\frac{2a}{y_1}.$

PICTURE. The lavender straight line touching the curve at $P$ is the tangent. Its steepness is $m_t=2a/y_1$: high up (small $y_1$) it is steep; low down (large $y_1$) it is gentle. (See Tangent and Normal to a Conic.)

• $m_t$ ::: slope of the tangent = "how much up per one across."
• $\frac{2a}{y_1}$ ::: that slope written using only $a$ and the height $y_1$.

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Step 4 — Name the two rays leaving the story at $P$

WHAT. Two directions meet at $P$:

1. the incoming horizontal ray — slope $0$ (flat, no rise),
2. the focal radius $PF$ — the straight segment from $P=(x_1,y_1)$ to the focus $F=(a,0)$.

WHY. The reflective property claims the bounced ray is the focal radius. So we must compare the tangent against both these lines and show it treats them fairly (equal angles).

PICTURE. Coral arrow = incoming horizontal ray. Mint arrow = focal radius $PF$ pointing from $P$ down to $F$. Lavender line = tangent sitting between them. The angle $\alpha$ (coral) sits between the incoming ray and the tangent; the angle $\beta$ (mint) sits between the tangent and the focal radius. Our whole goal: show $\alpha =\beta$.

The slope of the focal radius, term by term: $m_{PF}=\underset{\text{rise over run from }P\text{ to }F}{\underbrace{\frac{y_1-0}{x_1-a}}}=\frac{y_1}{x_1-a}.$

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Step 5 — Measure angle $\alpha$ (incoming ray vs tangent)

WHAT. Find the angle between two lines whose slopes we know: slope $0$ and slope $m_t$.

WHY this tool. To get the angle between two slopes we use the standard between-lines formula $\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|.$

HOW. Put $m_1=m_t=\frac{2a}{y_1}$ and $m_2=0$ (and recall $y_1>0$, so the ratio is already positive): $\tan \alpha =\left|\frac{\frac{2a}{y_1}-0}{1+\frac{2a}{y_1}\cdot 0}\right|=\frac{2a}{y_1}.$

PICTURE. Zoom on the coral wedge $\alpha$ between the flat incoming ray and the tilted tangent. Its "opening" is measured by $\tan \alpha =2a/y_1$.

• numerator $\frac{2a}{y_1}-0$ ::: the gap in steepness (tangent minus flat).
• denominator $1+\frac{2a}{y_1}\cdot 0=1$ ::: the "flat" line makes this term collapse to $1$.

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Step 6 — Measure angle $\beta$ (tangent vs focal radius) and watch the cancel

WHAT. Same between-lines formula, now between the tangent slope $m_t=\frac{2a}{y_1}$ and the focal-radius slope $m_{PF}=\frac{y_1}{x_1-a}$.

WHY. If this comes out equal to $\tan \alpha$, the tangent bisects the angle and reflection sends the ray through $F$. This is the whole proof in one line — so we push through carefully.

HOW (term by term, keeping the $|\cdots |$ honest).

=\left|\frac{2a(x_1-a)-y_1^2}{\,y_1(x_1-a)+2ay_1\,}\right|.$$ Now use the curve fact $y_1^2=4ax_1$ in the **numerator**: $$2a(x_1-a)-4ax_1=2ax_1-2a^2-4ax_1=-2a^2-2ax_1=-2a(x_1+a).$$ And factor the **denominator**: $$y_1(x_1-a)+2ay_1=y_1(x_1-a+2a)=y_1(x_1+a).$$ Divide, and keep the absolute-value bars until the very end: $$\tan\beta=\left|\frac{-2a(x_1+a)}{y_1(x_1+a)}\right| =\frac{2a\,|x_1+a|}{|y_1|\,|x_1+a|} =\frac{2a}{|y_1|}=\frac{2a}{y_1}.$$ The last equality uses our two grounded facts: $x_1+a>0$ (since $x_1\ge 0$ on the curve), so $|x_1+a|=x_1+a$ cancels cleanly top and bottom; and $y_1>0$ by our working assumption, so $|y_1|=y_1$. **PICTURE.** The figure shades the top factor $(x_1+a)$ and the bottom factor $(x_1+a)$ in the same colour so you *see* them cancel. Geometrically $x_1+a$ is exactly the distance $PM=PF$ (which is why it is guaranteed positive) — the "equal-distance" rule from Step 1 is what makes them identical, so they must cancel. ![[deepdives/dd-maths-3.4.03-d2-s06.png]] > [!formula] The equality falls out > $$\tan\alpha=\tan\beta=\frac{2a}{y_1}\quad\Longrightarrow\quad \boxed{\alpha=\beta}.$$ > The tangent gives the incoming ray and the focal radius **equal angles**. By the law of reflection, the horizontal ray reflects **straight through $F$**. ∎ --- ## Step 7 — The special (degenerate) cases — never skip these A drawing must cover *every* height $y_1$, including the ones where formulas seem to break. **Case A — the vertex, $y_1=0$.** The point $P$ sits at the origin, the tangent is **vertical**, and $m_t=\frac{2a}{0}$ is undefined. That is not a failure: a vertical tangent means the incoming horizontal ray hits the mirror *head-on* (perpendicular) and bounces **straight back** along the axis — which passes through $F$. Correct, just measured as a special limit. **Case B — $y_1<0$ (lower branch).** Our proof assumed $y_1>0$. Flip the whole picture over the axis. The slope $\frac{2a}{y_1}$ changes sign, but the between-lines formula is wrapped in absolute-value bars $|\cdots|$, so both $\tan\alpha$ and $\tan\beta$ still equal $\frac{2a}{|y_1|}$. The angles stay equal on **both** halves — the property is symmetric top and bottom, exactly as promised in the working assumption. **Case C — focal radius vertical (when $x_1=a$).** Take, for instance, the parabola $y^2=12x$ (so $4a=12$, $a=3$) and the point $P=(3,6)$: here $x_1=a=3$, so $m_{PF}=\frac{6}{0}$ is undefined — the focal radius points straight up. The between-lines formula can't ingest a vertical line directly, but we can read the angle off the picture. The tangent slope there is $m_t=\frac{2a}{y_1}=\frac{6}{6}=1$, so the tangent makes $45^\circ$ with the horizontal; a **vertical** line makes $90^\circ-45^\circ=45^\circ$ with that same tangent. So $\beta=45^\circ=\alpha$. The equality survives; only the *bookkeeping* changed. **PICTURE.** Three mini-panels: the head-on vertex bounce, the mirror-image upper/lower rays, and the vertical focal radius — each still showing $\alpha=\beta$. ![[deepdives/dd-maths-3.4.03-d2-s07.png]] > [!mistake] The trap these cases guard against > "The formula $\frac{2a}{y_1}$ blows up at $y_1=0$, so the property fails there." — No. A blown-up slope just means a **vertical tangent**, which is a perfectly good mirror. Always translate a divide-by-zero into its *picture* before declaring defeat. --- ## The one-picture summary Everything at once: horizontal rays rain in from the right (coral), each strikes the green parabola, and every single reflected ray (mint) funnels through the one coral dot $F$. The lavender tangent at each hit-point is the "fair referee" splitting $\alpha=\beta$. The dashed directrix on the left is the silent partner that forced the cancellation in Step 6. ![[deepdives/dd-maths-3.4.03-d2-s08.png]] > [!recall]- Feynman: the whole walkthrough in plain words > We drew a curve where every point is the same distance from a dot ($F$) and a line (the directrix). We shot a flat, horizontal ray at it and marked where it hits ($P$). At that spot the curved mirror acts flat — that flat line is the tangent, and we measured its steepness with a derivative, getting $2a/y_1$. Two directions leave $P$: the flat ray we shot in, and the line straight to $F$. We measured the angle each makes with the tangent using the "angle-between-two-slopes" tool (which is really just the tangent-subtraction rule). The first angle came out $2a/y_1$. The second, after using the curve's own equation, *also* came out $2a/y_1$ — because a factor $(x_1+a)$, which is really the equal-distance from Step 1 and is always positive, cancelled top and bottom. Equal angles means the mirror bounces the flat ray straight into $F$. And we checked the weird spots — the tip (vertical tangent, ray bounces back), the top vs bottom halves (mirror images), and the vertical focal radius (still $45^\circ=45^\circ$). Every ray, every case: into the focus. --- ## Active Recall > [!recall] Quick self-test > 1. Which single fact from Step 1 is the reason the $(x_1+a)$ factors cancel in Step 6? > 2. Why must we use a derivative in Step 3 rather than reading a slope off the equation directly? > 3. What does a "divide by zero" in $m_t=2a/y_1$ mean *geometrically*, and is the property still true there? > 4. On the lower branch $y_1<0$, why is $\alpha=\beta$ still guaranteed? Related bounces: [[Reflective property of ellipse (whispering galleries)]], [[Reflective property of hyperbola]], the optimisation view in [[Fermat's principle and shortest path]], and the tech in [[Applications of Conics in Engineering]]. The equal-distance fact that drove the cancel :::= $PF=PM$ Tangent slope of $y^2=4ax$ at $(x_1,y_1)$ ::: $\dfrac{2a}{y_1}$ Meaning of $m_t$ being undefined at the vertex ::: the tangent is vertical, so the ray reflects straight back along the axis (still through $F$) Why the factor $x_1+a$ may be cancelled without a sign worry ::: on $y^2=4ax$ we have $x_1\ge 0$, so $x_1+a>0$ always