How this instrument works
A focus is not the center — an ellipse has two of them, and they carry a strange defining property: pick any point on the boundary, add its distance to each focus, and the total is always the same constant, 2a. That single rule is why a loop of string looped around two pins traces a perfect ellipse as a pencil pulls it taut and slides around the loop — the string's length never changes, so the two distances it spans can't trade off in any other way.
The focal distance c falls out of a right triangle hiding inside the shape. Walk out from the center along the minor axis to where the curve crosses it: that point sits b units from center, and by the string-and-pins symmetry it must also sit exactly a units from each focus, since it splits the loop evenly between them. Center, focus, and that crossing point form a right triangle with legs b and c and hypotenuse a, so a² = b² + c² — the ordinary Pythagorean theorem, rearranged here to solve for c instead of the hypotenuse.
Close the gap between the two axis lengths and the foci creep toward the middle; make a and b equal and c drops to zero, both foci landing on the same spot — a circle is nothing more than the ellipse whose pair of foci has merged into one point. Pull the axes further apart instead and c climbs toward a, producing a long, narrow curve with its foci pushed out almost to the pointed ends.
- Type the ellipse's long half-axis length into the Semi-major axis, a field.
- Type the short half-axis length into the Semi-minor axis, b field — it has to stay at or under a.
- The Focal distance, c (from center) field updates at once, giving the distance from center out to each focus.
- Step off that same reading in both directions along the major axis to mark where the two actual focus points sit.
Worked example — an ellipsoidal reflector, a = 5, b = 3
A stage-lighting technician builds an ellipsoidal reflector whose cross-section has a semi-major axis a = 5 units and a semi-minor axis b = 3 units. The focal distance comes out to c = √(5² − 3²) = √(25 − 9) = √16 = 4 units exactly. The lamp filament is mounted at one focus, 4 units out from the reflector's center along the long axis; the shutter aperture sits at the other focus, 8 units away on the opposite side — twice c, since the two foci sit symmetrically either side of center.
That placement is not decorative. Because every point on the reflector's curve satisfies the same sum-of-distances rule, any ray of light leaving the filament and bouncing once off the elliptical wall arrives at the aperture having traveled a combined path of exactly 2a = 10 units, no matter which direction it left in — so the whole beam converges through that one small opening instead of scattering. Scaling the same design up to a = 13, b = 5 pushes the focal distance to c = √(169 − 25) = √144 = 12 units, the 5-12-13 right triangle turning up directly inside the geometry.
Questions
What exactly are the foci of an ellipse?
They are two fixed points inside the ellipse, each sitting c = √(a² − b²) units from the center along the major axis, with the property that the sum of the distances from any point on the boundary to both foci is always the same constant, 2a. That sum-to-a-constant rule is the actual definition of an ellipse; the familiar oval shape is just what results from graphing every point that satisfies it.
How is the formula c = √(a² − b²) derived?
From a right triangle hidden in the shape. The point where the ellipse crosses its minor axis sits b units from center and, by symmetry, exactly a units from each focus. Center, focus, and that crossing point form a right triangle with legs b and c and hypotenuse a, so a² = b² + c² — the Pythagorean theorem, solved here for c instead of the hypotenuse.
What happens to the foci when a and b are equal?
c = √(a² − a²) = 0, so both foci sit on top of the center point and the curve is a plain circle. A circle is the special, zero-eccentricity member of the ellipse family — the one case where the two axes match and the pair of foci has nowhere left to separate to.
Why do distances from any point on the ellipse to both foci always add to 2a?
That constant sum is the defining property of the curve, not a side effect of the formula — it's exactly how a loop of fixed-length string traces an ellipse around two pins. Checked at the vertex farthest from center, the two distances are a + c and a − c; add them and the c terms cancel, leaving 2a, the same total any other boundary point also gives.
What is eccentricity, and how does it connect to the foci?
Eccentricity, e = c / a, measures how far the foci sit from center relative to the ellipse's overall length: 0 for a circle, approaching 1 for a very flattened shape. In the worked reflector example above, e = 4 / 5 = 0.8, a noticeably elongated ellipse; a gently oval shape with b close to a instead gives an eccentricity close to 0.
Where does this focus geometry actually show up?
Two places worth knowing. Kepler's first law places the Sun at one focus, not the center, of every planet's elliptical orbit — the other focus sits empty in space. And in a whispering-gallery room built to an elliptical cross-section, sound leaving one focus reflects off the curved wall and converges exactly on the other, the same reflective property that steers light in an ellipsoidal lighting reflector.