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

Before we start, two words we lean on repeatedly:

Axis — the mirror line of the parabola. For $y^{2} = 4 a x$ it is the $x$ -axis. "Parallel to the axis" means a horizontal ray.
Focal radius $P F$ — the straight segment from the point $P$ on the curve to the focus $F = (a, 0)$ .

True or false — justify

A ray parallel to the axis always reflects through the focus, whatever point it hits

True — the equal-angle proof holds at every point

P

on the curve, so the hit location never matters.

A ray parallel to the axis but hitting the parabola from outside (convex side) still focuses

False — reflection needs the ray to strike the mirrored (concave) face; a ray on the back/convex side just glances off and does not obey the focusing geometry.

The reflective property depends on the wavelength or type of wave (light vs radio)

False — it is pure geometry (equal angles about the tangent), so it works identically for light, radio, sound or water ripples; only the mirror's smoothness matters in practice.

For a parabola, rays leaving the focus emerge parallel to the axis

True — reflection is reversible: run the parallel-in-to-focus path backwards and a source at the focus produces a parallel outgoing beam.

Rays leaving the focus of a parabola come back to the focus

False — that is an ellipse (Reflective property of ellipse (whispering galleries), focus-to-focus). A parabola sends focus rays out parallel; its "second focus" is at infinity.

A wider dish (larger rim) changes where the focus sits for the same value of

a

False — the focus is fixed at

(a, 0)

by the value of

a

alone; widening the rim gathers more energy but the focal spot stays put.

The property still holds if the ray comes in at a slight angle to the axis

False — only exactly parallel rays meet at the focus. A slightly tilted parallel beam converges near, but not exactly at, the focus (this off-axis blur is called coma in telescopes).

Equating angles with the tangent gives a different reflection law than equating with the normal

False — the normal is perpendicular to the tangent, so equal angles about one is automatically equal angles about the other; both give the identical reflected ray.

Spot the error

"In

y^{2} = 20 x

the focus is at

(20, 0)

because the coefficient is 20." — find the slip

Match

4 a = 20 \Rightarrow a = 5

; the focus is

(a, 0) = (5, 0)

, not

(4 a, 0)

. The

20

is the latus-rectum-length constant, not the focal distance.

"The tangent slope of

y^{2} = 4 a x

(x_{1}, y_{1})

\frac{2 a}{x _{1}}

." — fix it

Differentiating

2 y y^{'} = 4 a

gives

y^{'} = \frac{2 a}{y _{1}}

, denominator is

y_{1}

not

x_{1}

. The slope depends on the

y

-coordinate.

"Since

P F = P M

, the incoming ray must physically travel the distance

P M

before reflecting." — what's wrong

P F = P M

is a distance identity used to prove equal angles; it does not mean the light retraces the segment

P M

. The perpendicular to the directrix just encodes the ray's horizontal direction.

"The reflected ray bisects the angle at

P

." — correct the claim

It is the tangent that bisects the angle between the incoming ray and the focal radius, not the reflected ray. The reflected ray is the focal radius

P F

"A parabola has two foci, like an ellipse, so signals can focus at either one." — spot the trap

A parabola has exactly one focus. Its second focus has receded to infinity, which is precisely why the other family of rays comes out parallel instead of meeting a second point.

"At the vertex the tangent slope is

\frac{2 a}{0}

, so the property breaks there." — resolve it

At the vertex

y_{1} = 0

the tangent is vertical (slope undefined), and a horizontal ray hitting the vertex bounces straight back along the axis — which does pass through the focus. The property holds; only the slope formula fails, not the geometry.

"To find the focus of a dish, plug the rim width straight into

y = 4 a x

." — fix the setup

The equation is

y^{2} = 4 a x

(note the square) and you use the half-width as

y

with the depth as

x

, e.g. rim point

(x, y)

, then solve

y^{2} = 4 a x

for

a

Why questions

Why must the reflection be analysed about the tangent line rather than the whole curve

Reflection is a local event at the single point

P

; only the curve's instantaneous direction there — its tangent — governs the bounce, so we replace the curve by its tangent at

P

Why does the algebra factor

(x_{1} + a)

cancel in the proof of

tan α = tan β

The common factor top and bottom signals the balance built into the focus–directrix definition; its cancellation is exactly what forces

tan β = \frac{2 a}{y _{1}} = tan α

, proving equal angles.

Why do we place a bulb (not a lens) at the focus of a torch

Because a source at the focus sends every reflected ray out parallel to the axis, giving a tight beam — reversing the parallel-in-to-focus property. See Applications of Conics in Engineering.

Why is a parabola better than a spherical mirror for a telescope

A sphere only approximately focuses parallel rays (spherical aberration), while a parabola focuses all parallel rays exactly at one point, giving a sharp image.

Why is the reflective property an instance of Fermat's principle

Reflecting

F

across the tangent lands on the directrix, making the tangent the perpendicular bisector of

F M

; light travels the equal-angle path because it is the shortest (least-time) route — the heart of Fermat's principle and shortest path.

Why does widening the dish improve reception even though the focus doesn't move

A larger area intercepts more incoming parallel signal, and all of it is funnelled to the same focus, so the receiver collects more energy — greater gain, same focal point.

Edge cases

A horizontal ray strikes the vertex

(0, 0)

— where does it reflect

Straight back along the axis; since the axis passes through

F = (a, 0)

, it still reaches the focus, so the property survives the degenerate slope.

A ray hits the point where the parabola crosses the latus rectum,

P = (a, 2 a)

— is anything special

Tangent slope there is

\frac{2 a}{2 a} = 1

, and the focal radius

P F

is vertical; the tangent makes

4 5^{\circ}

with both — a clean symmetric check that equal angles hold.

What happens to the "focus" of the reflected family when we let the parabola flatten (

a \to \infty

)

The curve tends to a flat mirror and the focus recedes to infinity; parallel rays reflect to parallel rays — the focusing power vanishes smoothly, consistent with the limit.

A ray parallel to the axis but travelling away from the parabola (never hitting it) — does the property say anything

No; the property only concerns rays that actually strike the concave surface. A non-incident ray simply never reflects.

For the downward parabola

x^{2} = 4 a y

, does "parallel to the axis" still mean horizontal

No — its axis is the

y

-axis, so parallel rays are vertical, and they focus at

(0, a)

. Always identify the axis first before calling a ray "parallel."

Does the property hold for a point on the lower branch

y_{1} < 0

Yes; the proof used

y_{1}^{2} = 4 a x_{1}

, which is sign-independent, and

tan α = tan β = \frac{2 a}{y _{1}}

in magnitude — the lower half mirrors the upper half by symmetry about the axis.

Recall One-line reset

Parallel-in → focus; focus-out → parallel; equal angles about the tangent; single focus; $(a, 0)$ from $4 a$ . Everything else is a variation on these five.