How Parabolic Antennas Work

The mathematical proof that a parabola reflects all parallel rays to a single focus

The Principle of the Parabolic Reflector

This page shows mathematically why a parabolic reflector works. Parallel electromagnetic rays traveling along the parabola's axis of symmetry reflect to a single point — the focus — and conversely, waves emitted from a source placed at the focus reflect off the parabola and propagate outward as parallel rays.

Strictly speaking, the parabolic dish is a reflector. The antenna itself is the feed horn placed at the focus, which receives or emits the signal. The reflector shapes the beam.

Mathematical Derivation

Consider an electromagnetic ray traveling parallel to the parabola's axis (the y-axis) that strikes the inner surface of the parabola.

Incoming vertical ray striking a parabola at point M, reflecting to the focus on the y-axis
Figure 1. Geometry of an incoming electromagnetic ray reflecting off a parabola.

Proof: All Reflected Rays Pass Through the Focus

Let \( M(a,b) \) be the point where the ray strikes the parabola, \( i \) the angle of incidence with respect to the normal, and \( r \) the angle of reflection. By the law of reflection, \( i = r \).

Step 1 — Tangent and normal slopes

The standard equation of a parabola with vertex at the origin is:

\[ y = \frac{x^2}{4f} \]

Differentiating:

\[ y' = \frac{x}{2f} \]

So the tangent slope at \( M(a,b) \) is \( m_t = \dfrac{a}{2f} \), and the normal slope is:

\[ m_n = -\frac{2f}{a} \]

Step 2 — Relate angles to the normal

Let \( n \) be the angle the normal makes with the x-axis, so \( \tan(n) = -\dfrac{2f}{a} \).

The incoming ray is vertical, so the angle between the incoming ray and the normal equals the angle between the incoming direction (\(90°\) from the x-axis) and the normal direction. Call this \( i \). From the geometry, \( i = 90° - n \).

Step 3 — Slope of the reflected ray

The reflected ray makes an angle with the x-axis equal to the incoming direction minus the reflection deflection, which works out to \( 2n - 90° \). Hence:

\[ m_r = \tan(2n - 90°) = -\cot(2n) = -\frac{1 - \tan^2(n)}{2 \tan(n)} \]

Substituting \( \tan(n) = -\dfrac{2f}{a} \):

\[ m_r = -\frac{1 - \dfrac{4f^2}{a^2}}{-\dfrac{4f}{a}} = \frac{a^2 - 4f^2}{4af} \]

Step 4 — Show the reflected ray hits the y-axis at \( y = f \)

The reflected ray passes through \( M(a,b) \) with slope \( m_r \):

\[ y - b = m_r(x - a) \]

Set \( x = 0 \) to find the y-intercept:

\[ y_{\text{int}} = b - a\,m_r \]

Since \( M \) is on the parabola, \( b = \dfrac{a^2}{4f} \). Substituting:

\[ y_{\text{int}} = \frac{a^2}{4f} - a\cdot\frac{a^2 - 4f^2}{4af} = \frac{a^2 - a^2 + 4f^2}{4f} = \frac{4f^2}{4f} = f \]

Conclusion

The y-intercept of every reflected ray is \( f \), so every incoming parallel ray reflects through the focus \( (0, f) \) regardless of where it strikes the parabola.

The calculation was carried out for \( a > 0 \) (the right-hand side of the parabola). The case \( a < 0 \) follows by the parabola's mirror symmetry across the y-axis. This one derivation therefore covers every possible incoming parallel ray.

Modes of Operation

Mode Rays Where energy is concentrated
Receiving Parallel rays arriving from space All converge at the focus, where the receiver sits
Emitting Rays emanating from a source at the focus Reflect into a narrow parallel beam toward the target

Receiving Mode

Parallel incoming rays reflected by a parabolic dish and converging at its focus
Incoming parallel rays (e.g. satellite signals) reflect off the dish and converge precisely at the focus, where the receiver feed is located.

Emitting Mode

Rays emanating from the focus of a parabolic dish and reflecting outward as parallel lines
Rays emitted from a source at the focus reflect off the dish and travel outward as a tight parallel beam.

At radio wavelengths, diffraction causes some energy to spread slightly beyond the ideal focal point. The geometric-optics derivation above assumes wavelengths small enough that ray behavior dominates — a good approximation for satellite, WiFi, and microwave dishes, but not a perfect one.

Practice Exercises

Find the focus of each parabola. Expand View Solution to check your work.

Exercise 1 — \( y = 4x^2 \)

View Solution

Match against \( y = \dfrac{1}{4f} x^2 \):

\[ \frac{1}{4f} = 4 \;\Rightarrow\; 1 = 16f \;\Rightarrow\; f = \frac{1}{16} \]

Focus: \( \left(0, \tfrac{1}{16}\right) \)

Exercise 2 — \( y = \dfrac{x^2}{8} \)

View Solution

Rewrite as \( y = \dfrac{1}{8} x^2 \) and match:

\[ \frac{1}{4f} = \frac{1}{8} \;\Rightarrow\; 4f = 8 \;\Rightarrow\; f = 2 \]

Focus: \( (0, 2) \)

Exercise 3 — \( y = 0.12 x^2 \)

View Solution

Match against \( \dfrac{1}{4f} = 0.12 \):

\[ 1 = 0.48f \;\Rightarrow\; f = \frac{1}{0.48} = \frac{25}{12} \approx 2.083 \]

Focus: \( \left(0, \tfrac{25}{12}\right) \)

Frequently Asked Questions

Why do parabolic antennas use a parabola?

A parabola is the unique curve that reflects every parallel incoming ray exactly to a single focus, and conversely radiates every ray from the focus as a perfect parallel beam. That makes it ideal for focusing weak signals and forming narrow directional beams.

What is the focal distance of a parabolic dish?

For \( y = \dfrac{x^2}{4f} \), the focal distance is \( f \) — the distance from the vertex to the focus along the axis. For a dish with diameter \( D \) and depth \( d \), \( f = \dfrac{D^2}{16d} \).

What happens if the feed is not at the focus?

If the feed is off the focus, the reflected rays no longer converge on the feed. This causes signal loss, beam distortion, and reduced efficiency — which is why dish alignment and feed positioning are critical in real installations.

Is the parabolic dish itself the antenna?

Strictly, no — the dish is a reflector. The actual antenna is the feed horn placed at the focus, which receives or emits the signal. The reflector shapes the beam and concentrates the energy.

Further References on Parabolic Reflectors

Last updated: October 17, 2026