Reflective property of parabola (application in telescopes, antennas): Exercises

A reminder of the two symbols we use everywhere:

The first figure below is your map for every problem on this page. Read it slowly:

the blue curve is the parabola $y^{2} = 4 a x$ ;
the green arrow is the incoming ray, arriving horizontally (parallel to the axis);
the yellow dashed line is the tangent at $P$ — the mirror surface at that point;
the red arrow is the focal radius $P F$ , running from $P$ to the yellow focus $F$ ;
the green $α$ and red $β$ mark the two angles the tangent makes — the picture shows them equal.

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

Level 1 — Recognition

L1·Q1

State, in your own words, the reflective property of a parabola. Then: for $y^{2} = 20 x$ , where is the focus?

Recall Solution

Property (one sentence): any ray travelling parallel to the axis reflects off the curve and passes through the single focus; equivalently the tangent makes equal angles ( $α = β$ ) with the incoming ray and the focal radius.

Focus: compare $y^{2} = 20 x$ with $y^{2} = 4 a x$ . Match $4 a = 20 \Rightarrow a = 5$ . Why $a$ and not $4 a$ ? The focus is at $(a, 0)$ , so we must peel the $4$ off first. Focus $= (5, 0)$ .

L1·Q2

For $y^{2} = 8 x$ , write down (a) the value of $a$ , (b) the focus, (c) the directrix, and (d) the tangent slope at the point $(2, 4)$ .

Recall Solution

(a) $4 a = 8 \Rightarrow a = 2$ . (b) Focus $= (a, 0) = (2, 0)$ . (c) Directrix is the vertical line $x = - a$ , i.e. $x = - 2$ . (d) Tangent slope $= \frac{2 a}{y _{1}} = \frac{2 \cdot 2}{4} = 1$ . Check $(2, 4)$ is on the curve: $4^{2} = 16 = 8 \cdot 2$ ✓.

Level 2 — Application

L2·Q1

A parabolic mirror has equation $y^{2} = 16 x$ . A horizontal ray hits it at $P = (4, 8)$ . Through which point does the reflected ray pass?

Recall Solution

Step 1 — find $a$ . $4 a = 16 \Rightarrow a = 4$ . Why? We need the focus. Step 2 — focus. $F = (a, 0) = (4, 0)$ . Step 3 — check $P$ on curve. $8^{2} = 64 = 16 \cdot 4$ ✓. Step 4 — apply the property. A ray parallel to the axis reflects through the focus, so it passes through $(4, 0)$ . Sanity: $P = (4, 8)$ and $F = (4, 0)$ share the same $x$ , so this particular ray reflects straight down the vertical line $x = 4$ into the focus.

L2·Q2

Design problem. A satellite dish has cross-section $y^{2} = 4 a x$ . It is $4$ m wide at a depth of $1$ m. How far from the vertex must the receiver sit?

Recall Solution

Step 1 — read the rim point. " $4$ m wide" means half-width $y = 2$ ; "depth $1$ m" means $x = 1$ . So the rim point is $(1, 2)$ . Step 2 — plug in. $y^{2} = 4 a x \Rightarrow 2^{2} = 4 a (1) \Rightarrow 4 = 4 a \Rightarrow a = 1$ . Step 3 — locate the receiver. The receiver goes at the focus $(a, 0) = (1, 0)$ , i.e. $1$ m from the vertex along the axis. Why: the reflective property guarantees all incoming parallel waves converge there.

L2·Q3

Verify the equal-angle property numerically for $y^{2} = 16 x$ at $P = (4, 8)$ (so $a = 4$ ). Compute $tan α$ and $tan β$ .

Recall Solution

Tangent slope: $m_{t} = \frac{2 a}{y _{1}} = \frac{8}{8} = 1$ . $tan α$ (horizontal slope $0$ vs tangent slope $1$ ): $tan α = \frac{1 - 0}{1 + 0} = 1 \Rightarrow α = 4 5^{\circ} .$ Focal radius slope: $m_{P F} = \frac{8 - 0}{4 - 4}$ = vertical (undefined). Why a vertical line makes $4 5^{\circ}$ with a slope- $1$ line. The angle-between-lines formula $tan θ = \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}}$ blows up when a slope is infinite, so we take the limit as $m_{2} \to \infty$ (the focal radius becoming vertical). Divide top and bottom by $m_{2}$ : $tan β = \frac{m _{1} / m _{2} - 1}{1/ m _{2} + m _{1}} m_{2} \to \infty \frac{0 - 1}{0 + m _{1}} = \frac{1}{∣ m _{1} ∣} = \frac{1}{1} = 1.$ So $β = 4 5^{\circ}$ . (Geometrically: a vertical line is $9 0^{\circ}$ from horizontal; the slope- $1$ tangent is $4 5^{\circ}$ from horizontal; their difference is $9 0^{\circ} - 4 5^{\circ} = 4 5^{\circ}$ .) $α = β = 4 5^{\circ}$ ✓ — the property holds here.

Level 3 — Analysis

L3·Q1

For a general point $P = (x_{1}, y_{1})$ on $y^{2} = 4 a x$ with $y_{1} > 0$ , prove from scratch that $tan α = tan β = \frac{2 a}{y _{1}}$ . State each algebraic step's purpose.

Recall Solution

Tools we need and why. The reflection happens about the tangent, so we need the tangent slope; we compare it against the incoming ray (slope $0$ ) and the focal radius (slope $m_{P F}$ ) using the angle-between-lines formula $tan θ = \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}} .$ Here $m_{1}$ and $m_{2}$ are simply the slopes of the two lines whose angle we want, and $θ$ is the angle between them. Why this formula? It answers exactly "what is the angle between two lines given their slopes?" — which is precisely $α$ and $β$ .

Tangent slope. Differentiate $y^{2} = 4 a x$ implicitly: $2 y y^{'} = 4 a \Rightarrow m_{t} = \frac{2 a}{y _{1}}$ .

Angle $α$ (horizontal slope $0$ vs $m_{t}$ ): $tan α = \frac{\frac{2 a}{y _{1}} - 0}{1 + 0} = \frac{2 a}{y _{1}} .$

Angle $β$ (tangent $m_{t}$ vs $m_{P F} = \frac{y _{1}}{x _{1} - a}$ ):

= \left|\frac{2a(x_1-a) - y_1^2}{y_1(x_1-a) + 2a y_1}\right|.$$ **Numerator** (substitute $y_1^2 = 4ax_1$, *because $P$ is on the curve*): $$2a(x_1-a) - 4ax_1 = 2ax_1 - 2a^2 - 4ax_1 = -2a(x_1 + a).$$ **Denominator:** $y_1(x_1 - a + 2a) = y_1(x_1 + a).$ $$\tan\beta = \left|\frac{-2a(x_1+a)}{y_1(x_1+a)}\right| = \frac{2a}{y_1}.$$ The factor $(x_1+a)$ cancels top and bottom, leaving $\tan\alpha = \tan\beta = \dfrac{2a}{y_1}$, so $\alpha = \beta$. ∎

L3·Q2

Take the same parabola and reflect the focus $F = (a, 0)$ across the tangent line at $P$ . Show geometrically that the reflected image lands on the directrix, and explain why this re-proves the reflective property. (Refer to the figure.)

Recall Solution

Let $M$ be the foot of the perpendicular from $P$ to the directrix $x = - a$ . By the focus–directrix definition of a parabola, $P F = P M$ . Note that $M$ is exactly where the horizontal incoming ray points: the ray travels toward the directrix, hitting it at $M$ .

Since $α = β$ (proved in L3·Q1), the tangent bisects the angle $∠ F P M$ (the angle at $P$ between the focal radius $P F$ and the ray direction $P M$ ). With $P F = P M$ and the tangent bisecting $∠ F P M$ , the tangent is the perpendicular bisector of the segment $F M$ . Reflecting $F$ across the tangent therefore lands exactly on $M$ , a point on the directrix.

Why this re-proves the property: a horizontal incoming ray heads toward $M$ . Reflecting across the tangent sends "toward $M$ " to "toward the mirror image of $M$ ", which is $F$ . So the reflected ray goes straight to the focus. See the second figure: the tangent is the mirror, $F$ and $M$ are mirror-images.

This is the Fermat / shortest-path view in disguise — light takes the equal-angle (extremal) path.

Level 4 — Synthesis

L4·Q1

The point $P$ on $y^{2} = 4 a x$ at the end of the latus rectum is $(a, 2 a)$ . Show the reflected ray there makes exactly $4 5^{\circ}$ with the axis, for every parabola (any $a > 0$ ).

Recall Solution

Step 1 — tangent slope at $(a, 2 a)$ . $m_{t} = \frac{2 a}{y _{1}} = \frac{2 a}{2 a} = 1$ . Independent of $a$ . Step 2 — incoming ray angle. Incoming slope $0$ vs tangent slope $1$ : $tan α = \frac{1 - 0}{1} = 1 \Rightarrow α = 4 5^{\circ}$ . Step 3 — reflected ray direction. The reflected ray goes to the focus $F = (a, 0)$ ; from $P = (a, 2 a)$ this is the vertical line $x = a$ , which makes $9 0^{\circ}$ with the axis — but its angle with the tangent is again $4 5^{\circ}$ ( $β = 4 5^{\circ}$ ), confirming $α = β$ . Conclusion: at the latus-rectum tip the tangent is always inclined at $4 5^{\circ}$ and the reflected ray drops vertically into the focus — for any $a$ . The shape scales but the angle is universal.

L4·Q2

A car headlight reflector is $y^{2} = 4 a x$ with the bulb at the focus. The reflector's rim is at $x = 9$ cm and the focus sits $4$ cm from the vertex. (a) Find $a$ and the equation. (b) Find the rim's half-height. (c) Explain, using the reflective property run backwards, why the emitted beam is parallel.

Recall Solution

(a) Focus at $(a, 0)$ is $4$ cm from vertex $\Rightarrow a = 4$ . Equation: $y^{2} = 16 x$ . (b) At the rim $x = 9$ : $y^{2} = 16 \cdot 9 = 144 \Rightarrow y = 12$ . Half-height $= 12$ cm (full height $24$ cm). (c) The property says parallel-in $\to$ focus. Light is reversible, so focus-out $\to$ parallel: a bulb at the focus sends every ray to the curve, where it reflects to travel parallel to the axis. That is the "PIF–FOP" mnemonic's second half — a searchlight beam.

Level 5 — Mastery

L5·Q1

An incoming horizontal ray strikes $y^{2} = 4 x$ at $P = (4, 4)$ . Find the equation of the reflected ray (the line from $P$ through the focus), then find where this reflected ray crosses the $x$ -axis. Confirm it is the focus.

Recall Solution

Step 1 — $a$ and focus. $4 a = 4 \Rightarrow a = 1$ ; focus $F = (1, 0)$ . Step 2 — check $P$ . $4^{2} = 16 = 4 \cdot 4$ ✓. Step 3 — line $P F$ . Slope $m_{P F} = \frac{4 - 0}{4 - 1} = \frac{4}{3}$ . Equation through $F = (1, 0)$ : $y - 0 = \frac{4}{3} (x - 1)$ , i.e. $y = \frac{4}{3} (x - 1)$ or $4 x - 3 y - 4 = 0$ . Step 4 — cross the $x$ -axis. Set $y = 0$ : $\frac{4}{3} (x - 1) = 0 \Rightarrow x = 1$ . Crossing point $(1, 0) = F$ ✓. The reflected ray indeed passes through the focus.

L5·Q2

Two-ray convergence. Two horizontal rays hit $y^{2} = 4 x$ at $P_{1} = (4, 4)$ and $P_{2} = (1, - 2)$ . Find both reflected-ray lines and prove they intersect exactly at the focus.

Recall Solution

Here $a = 1$ , focus $F = (1, 0)$ . Ray 1 at $P_{1} = (4, 4)$ : slope $\frac{4 - 0}{4 - 1} = \frac{4}{3}$ , line $y = \frac{4}{3} (x - 1)$ . Ray 2 at $P_{2} = (1, - 2)$ : check on curve $(- 2)^{2} = 4 = 4 \cdot 1$ ✓. Slope $\frac{- 2 - 0}{1 - 1}$ = vertical, line $x = 1$ . Intersection. Substitute $x = 1$ into Ray 1: $y = \frac{4}{3} (1 - 1) = 0$ . Point $(1, 0) = F$ . Both reflected rays pass through $F$ , so they meet at the focus — exactly the collecting behaviour that makes a dish work. ∎

L5·Q3

Degenerate / boundary case — the vertex. What happens to a horizontal ray that arrives along the axis itself and hits the vertex $(0, 0)$ ? Discuss why the tangent-slope formula "breaks" and what physically occurs.

Recall Solution

At the vertex $y_{1} = 0$ , so $m_{t} = \frac{2 a}{y _{1}} = \frac{2 a}{0}$ is undefined — the tangent there is vertical (the $y$ -axis). Why the formula breaks: division by $y_{1} = 0$ signals a vertical tangent, not a failure of the geometry. A ray coming in exactly along the axis is the horizontal $x$ -axis itself. It strikes the vertical tangent head-on (perpendicular, $α = 9 0^{\circ}$ ) and reflects straight back along the axis. Since the axis already passes through the focus $(a, 0)$ , the reflected ray still goes through the focus — the property survives even in this limiting case, we just read the tangent as vertical. Boundary conclusion: the axis-ray is the one ray that is its own focal radius; equal angles ( $9 0^{\circ} = 9 0^{\circ}$ ) still hold.

Active Recall

For deeper connections see Reflective property of ellipse (whispering galleries), Reflective property of hyperbola, and Applications of Conics in Engineering.

Reflective property in one line

a ray parallel to the axis reflects through the focus; the tangent makes equal angles

α = β

with the incoming ray and the focal radius

Focus of

y^{2} = 20 x

(5, 0)

since

4 a = 20 \Rightarrow a = 5

Receiver for a dish

4

m wide at depth

1

at the focus

(1, 0)

, i.e.

1

m from the vertex

Tangent slope at latus-rectum tip

(a, 2 a)

1

(the

a

's cancel — all parabolas are similar)

Ray hitting the vertex

reflects straight back along the axis, still through the focus (tangent is vertical there)