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

This page is the drill room for the parent note. There we proved that a ray parallel to the axis reflects through the focus. Here we hit every kind of case the topic can throw at you — different quadrants, the degenerate ray, the vertex, a downward-opening dish, a real word problem, and an exam trap.

Before starting, three symbols we will lean on (defined so a newcomer is never lost):

The scenario matrix

Every parabola-reflection problem is one of these cells. The examples below are labelled by cell.

Cell	What changes	Danger it tests
C1	$P$ in quadrant I ( $y_{1} > 0$ )	baseline, find focus point
C2	$P$ in quadrant IV ( $y_{1} < 0$ )	sign of tangent slope $\frac{2 a}{y _{1}}$ flips
C3	$P$ at the vertex $(0, 0)$	$y_{1} = 0$ → tangent slope undefined, ray behaviour
C4	Downward / sideways-opening ( $x^{2} = 4 a y$ , negative $a$ )	axis is vertical, focus flips sign
C5	Numeric equal-angle check	verify $α = β$ with real numbers
C6	Word problem (dish/torch design)	translate "width & depth" → find $a$
C7	Limiting case — ray far from axis / flat rim	does it still hit the focus?
C8	Exam twist — reversed (focus OUT) + wrong-focus trap	conceptual, no brute force

Examples

Worked example

C1 — Quadrant I, find the focus point Mirror $y^{2} = 8 x$ . A horizontal ray strikes at $P = (2, 4)$ . Where does it go after bouncing?

Forecast: Guess — does the reflected ray hit a point, or shoot off? To what point?

Match $y^{2} = 8 x$ with $y^{2} = 4 a x \Rightarrow 4 a = 8 \Rightarrow a = 2$ . Why this step? The focus of a standard parabola is $(a, 0)$ ; we cannot name it until we know $a$ .
So $F = (2, 0)$ . Why this step? Direct read-off from the standard form.
Check $P$ is really on the curve: $4^{2} = 16$ and $8 \cdot 2 = 16$ ✓. Why this step? The reflective property only applies to points on the parabola.
By the reflective property, the ray reflects through $F = (2, 0)$ . Why this step? This is the payoff we were building toward — once $F$ is pinned and $P$ is confirmed on the curve, the property guarantees the reflected ray lands on $F$ , so no further computation is needed.

Verify: In the figure the orange incoming ray runs horizontally into $P = (2, 4)$ ; the plum reflected arrow then drops straight down to $F = (2, 0)$ . Because $P$ and $F$ share $x = 2$ , that reflected segment is exactly vertical — a clean visual confirmation of the answer.

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

Worked example

C2 — Quadrant IV, the slope sign flips Same mirror $y^{2} = 8 x$ ( $a = 2$ ), but the ray hits the lower half at $P = (2, - 4)$ . Show it still goes to $F$ , and watch the tangent slope's sign.

Forecast: Will the tangent slope be $+ 1$ like before, or $- 1$ ? Does the ray still reach $(2, 0)$ ?

Tangent slope $m_{t} = \frac{2 a}{y _{1}} = \frac{4}{- 4} = - 1$ . Why this step? With $y_{1} < 0$ the ratio $\frac{2 a}{y _{1}}$ turns negative — the curve tilts the other way in the lower half.
Focal radius slope $m_{P F} = \frac{- 4 - 0}{2 - 2}$ = vertical (undefined). Why this step? $P$ and $F$ again share $x = 2$ , so $P F$ is a vertical line.
Angle of tangent (slope $- 1$ ) with a vertical line: a slope- $- 1$ line makes $4 5^{\circ}$ with the vertical, and the horizontal ray also makes $4 5^{\circ}$ with it. Why this step? Equal angles $\Rightarrow$ the reflection law is satisfied — the property is mirror-symmetric in $y$ .
Reflected ray passes through $F = (2, 0)$ . Why this step? Equal incidence and reflection angles about the tangent is the reflective property; confirming $α = β$ in step 3 is precisely what licenses us to conclude the ray reaches $F$ .

Verify: In the figure the dashed tangent through $P = (2, - 4)$ visibly slopes down-to-the-right (slope $- 1$ ), the mirror image of C1's up-slope. The plum arrow again climbs vertically to $F = (2, 0)$ , so the lower ray lands on the same focus. ✓

Worked example

C3 — The vertex, $y_{1} = 0$ (degenerate tangent slope) What happens to a ray that would hit at the vertex $(0, 0)$ of $y^{2} = 8 x$ ? The formula $m_{t} = \frac{2 a}{y _{1}}$ blows up (division by zero). Resolve it.

Forecast: Is the tangent flat, vertical, or nonexistent? Where does a vertex ray go?

Approach the vertex from both sides: let $P = (x_{1}, y_{1})$ with $y_{1} \to 0^{+}$ (upper branch), then $\frac{2 a}{y _{1}} \to + \infty$ ; and with $y_{1} \to 0^{-}$ (lower branch), $\frac{2 a}{y _{1}} \to - \infty$ . Why this step? A tangent is only well-defined if the slope agrees from both directions of approach — we must check $0^{+}$ and $0^{-}$ , not just one.
Both a huge $+ \infty$ slope and a huge $- \infty$ slope describe the same vertical line $x = 0$ (the $y$ -axis); a line becoming infinitely steep from either side stands straight up. So the tangent at the vertex is the vertical line $x = 0$ . Why this step? The two-sided limits agreeing on "vertical" is what makes the vertical tangent legitimate rather than a formula glitch.
A horizontal ray hitting a vertical mirror reflects straight back on itself (angle of incidence $0^{\circ}$ ). Why this step? Law of reflection: hitting a mirror head-on sends you straight back.
The reflected ray travels back along the $x$ -axis, and $F = (2, 0)$ lies on the $x$ -axis, so it passes through $F$ . Why this step? This closes the case: even the degenerate vertex ray obeys "parallel-in → through the focus," because the axis ray both enters and returns along the axis on which $F$ sits — the property survives the limit with no exception.

Verify: In the figure the tangent is drawn as the vertical dashed line $x = 0$ through the vertex; the orange axis ray comes in and the plum arrow returns straight along the $x$ -axis through $F = (2, 0)$ . Both the $0^{+}$ and $0^{-}$ limits point the tangent the same way. ✓

Worked example

C4 — Vertical axis, downward-opening dish A torch reflector has cross-section $x^{2} = - 8 y$ (opens downward). Find its focus, and state where the bulb goes for a parallel beam.

Forecast: For $x^{2} = 4 a y$ , is the focus on the $x$ -axis or the $y$ -axis? Above or below the vertex?

Match $x^{2} = - 8 y$ with $x^{2} = 4 a y \Rightarrow 4 a = - 8 \Rightarrow a = - 2$ . Why this step? Here the axis is vertical ( $y$ -axis), and $a$ carries the opening direction. Negative $a$ ⇒ opens downward.
For $x^{2} = 4 a y$ the focus is $(0, a) = (0, - 2)$ . Why this step? Swap the roles of $x$ and $y$ from the standard $y^{2} = 4 a x$ ; the focus sits on the axis of symmetry, here the $y$ -axis.
The reflective property now says: rays parallel to the $y$ -axis reflect through $(0, - 2)$ . Why this step? The property is about the axis, whichever direction it points.
Reversed (torch): put the bulb at the focus $(0, - 2)$ and light shoots out as a beam parallel to the $y$ -axis. Why this step? The reflection law works both directions — "Focus OUT → Parallel" is the reverse of "Parallel IN → Focus," so this final step converts our found focus into the actual engineering instruction for the torch.

Verify: $a = - 2$ has magnitude $2$ , so the bulb sits $2$ units from the vertex inside the bowl — matches physical torches. ✓

Worked example

C5 — Numeric equal-angle check ( $α = β$ ) For $y^{2} = 8 x$ ( $a = 2$ ) at $P = (2, 4)$ , compute $tan α$ (horizontal ray vs tangent) and $tan β$ (tangent vs focal radius) and confirm they match.

Forecast: Both should equal $\frac{2 a}{y _{1}}$ . Predict the number before computing.

Tangent slope $m_{t} = \frac{2 a}{y _{1}} = \frac{4}{4} = 1$ . Why this step? This is the reflecting line; both angles are measured from it.
$tan α = \frac{m _{t} - 0}{1 + m _{t} \cdot 0} = \frac{1}{1} = 1$ , so $α = 4 5^{\circ}$ . Why this step? Incoming ray is horizontal (slope $0$ ); the angle formula $tan θ = \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}}$ gives the wedge between two slopes.
Focal radius $P F$ : $m_{P F} = \frac{4 - 0}{2 - 2}$ = vertical. A vertical line vs the slope- $1$ tangent makes $4 5^{\circ}$ , so $β = 4 5^{\circ}$ . Why this step? We need $β$ to compare; a vertical direction is $9 0^{\circ}$ , tangent is $4 5^{\circ}$ , difference $= 4 5^{\circ}$ .
$tan α = tan β = 1 \Rightarrow α = β = 4 5^{\circ}$ . Why this step? Equating the two computed tangents is the whole point of the exercise: it numerically demonstrates the equal-angle claim the general proof made symbolically, so the reflection law is confirmed at this concrete point.

Verify: Both angles equal $4 5^{\circ}$ — the tangent is the fair referee, exactly as the general proof promised. ✓

Worked example

C6 — Word problem: design a satellite dish A dish's cross-section is $y^{2} = 4 a x$ . It measures $3$ m across at a depth of $0.5$ m. Where do you mount the receiver?

Forecast: Guess whether the focus is more or less than $0.5$ m deep — i.e. inside or outside the bowl.

" $3$ m across" ⇒ half-width $y = 1.5$ ; "depth $0.5$ m" ⇒ $x = 0.5$ . Rim point $= (0.5, 1.5)$ . Why this step? The rim is the farthest point; its coordinates pin down the shape.
Plug into $y^{2} = 4 a x$ : $1. 5^{2} = 4 a (0.5) \Rightarrow 2.25 = 2 a \Rightarrow a = 1.125$ . Why this step? One point on the curve determines the single unknown $a$ .
Focus $= (a, 0) = (1.125, 0)$ : mount the receiver $1.125$ m from the vertex along the axis. Why this step? All incoming parallel satellite waves converge there.
Note $1.125 > 0.5$ : the focus sits beyond the rim depth, so the receiver hangs above the bowl — physically correct for a shallow dish. Why this step? This is a physical-plausibility check that also answers the forecast: because $1.125 > 0.5$ the mounting point is outside the dish, exactly how real shallow dishes suspend their LNB on an arm.

Verify: Check the rim lies on the curve: $4 (1.125) (0.5) = 2.25 = 1. 5^{2}$ ✓. Units: metres throughout. ✓

Worked example

C7 — Limiting case: a ray near the flat rim On $y^{2} = 8 x$ ( $a = 2$ ), take a ray hitting very high up at $P = (50, 20)$ . Does it still focus at $(2, 0)$ , even though $P$ is far off-axis?

Forecast: Intuition may whisper "so far out, it must miss the focus." Test it.

Check on curve: $2 0^{2} = 400$ , $8 \cdot 50 = 400$ ✓. Why this step? Only genuine curve points obey the property.
Tangent slope $m_{t} = \frac{2 a}{y _{1}} = \frac{4}{20} = 0.2$ — a nearly flat tangent. Why this step? Far from the vertex the curve flattens; small slope is expected.
Focal radius slope $m_{P F} = \frac{20 - 0}{50 - 2} = \frac{20}{48} = \frac{5}{12} \approx 0.4167$ . Why this step? We compare its angle with the tangent to the incoming-ray angle.
$tan α = \frac{0.2 - 0}{1} = 0.2$ . $tan β = \frac{0.2 - \frac{5}{12}}{1 + 0.2 \cdot \frac{5}{12}} = \frac{- 0.21 6}{1.08 3} = 0.2$ . Why this step? Equal tangents ⇒ equal angles ⇒ the reflected ray hits $F = (2, 0)$ no matter how far off-axis — this is what refutes the forecast's worry.

Verify: $tan α = tan β = 0.2$ despite the extreme point — the property has no limit of validity on the curve. ✓

Worked example

C8 — Exam twist: reverse direction + wrong-focus trap "A telescope mirror is $y^{2} = 20 x$ . A student places the detector at $(5, 0)$ and claims light from the detector will leave as a parallel beam. A friend says the detector should be at $(20, 0)$ . Who is right, and what actually happens with reversed rays?"

Forecast: Which point is the true focus — $5$ or $20$ ? And does focus-out really give parallel?

Match $y^{2} = 20 x$ with $y^{2} = 4 a x \Rightarrow 4 a = 20 \Rightarrow a = 5$ . Why this step? $4 a$ is the coefficient, not the focal distance — the classic trap.
Focus $= (a, 0) = (5, 0)$ . The student is right; the friend confused $4 a = 20$ with the focus. Why this step? $(20, 0)$ would use $4 a$ as the coordinate — a wrong-focus error (parent Mistake C).
Reversed rays: by the property "Focus OUT → Parallel," a source at $(5, 0)$ emits rays that reflect parallel to the axis. Why this step? The reflection law is reversible — swapping arrow directions preserves equal angles.
So the telescope-in-reverse becomes a searchlight; the detector position is correct at $(5, 0)$ . Why this step? This states the verdict the question asked for — it ties the correct focus ( $a = 5$ ) to the correct physical behaviour (parallel emission), settling both the trap and the reversal in one line.

Verify: $4 a = 20 \Rightarrow a = 5$ , focus $(5, 0)$ , not $(20, 0)$ . The friend's $20$ is $4 a$ , off by a factor of $4$ . ✓

Recall Rapid self-test across the matrix

Q1. Which cell tests division by zero? A1. C3 — the vertex, where $y_{1} = 0$ makes $\frac{2 a}{y _{1}}$ blow up.

Q2. For $x^{2} = - 8 y$ , where is the focus? A2. $(0, - 2)$ — vertical axis, opens downward.

Q3. What is the focus of $y^{2} = 20 x$ ? A3. $(5, 0)$ , since $4 a = 20 \Rightarrow a = 5$ .

Q4. Does a far-off-axis ray still hit the focus? A4. Yes (C7) — the property holds at every point on the curve.

For neighbouring conics see Reflective property of ellipse (whispering galleries) and Reflective property of hyperbola; for the deeper "why" see Fermat's principle and shortest path and Applications of Conics in Engineering. The base curve is Parabola - standard equation y^2=4ax.