Derive the ellipse from its two foci

Try

Put two fixed points at F₁=(−3,0) and F₂=(3,0). An ellipse consists of points P whose distances to these foci add to 10. The point (0,4) works because each distance is 5. The point (5,0) also works because its distances are 8 and 2.

The equation x²/25+y²/16=1 describes the same curve, but that equivalence needs proof. Substituting a few successful points cannot rule out extra points introduced by algebra. We will derive the equation from the distance condition and then prove that every point of the equation satisfies the original condition.

Predict

The focal separation is 6. Predict why a distance sum of 4 is impossible. Any path from one focus to P and then to the other focus has length at least their direct separation, by triangle inequality. A sum equal to 6 gives the segment between the foci, not an ordinary ellipse. A nondegenerate ellipse needs a larger sum.

Use general foci (−c,0),(c,0), with c>0, and distance sum 2a, where a>c. The horizontal endpoints are (±a,0). At x=0 both distances are a, giving y²=a²−c². Define b=√(a²−c²)>0. These endpoints predict horizontal semiaxis a and vertical semiaxis b, but endpoints alone still do not determine the whole curve.

Experiment

Use the explorer to choose a and c, then predict the two focal distances and their sum at a probe. Test the right vertex and the upper vertex. Move the probe away from the curve and identify which condition fails. The live values may differ from the fixed a=5,c=3 example.

Test an ellipse against its focal definition

F1=(−c,0) and F2=(c,0). The positive target is 2a. An ellipse requires PF1+PF2=2a. Choose a point P and predict whether it meets that condition. The equation check is a separate representation, not a replacement for the focal definition.

Positive parameter a2345Focus distance c123456Reset focal explorer

a>c>0, so the ellipse is nondegenerate and b²=a²−c². Here b²=16.

Probe xProbe yPlace P at right vertexPlace P at upper vertexReflect P across y-axisReveal focal checks

The solid curve is the focal locus, drawn with bounded samples and clipped to this fixed window. Hyperbola branches and degenerate rays continue beyond it. Dashed PF1 and dotted PF2 appear after revealing; the square and open-circle focus markers supplement color.

Predict the two distances and decide which combination must stay fixed. Reveal the calculations after committing to a prediction.

Try the right and upper vertices, then move P one unit away and compare the distance sum. Change a and c to an equality case and an impossible case. Explain those cases using the triangle inequality before relying on a graph. Positive c keeps the two foci distinct in this explorer.

Coordinates and distances use floating-point arithmetic. Each control change hides old checks; this temporary exploration does not save a response or award mastery. Enter your derivation and its reverse implication in the written practice.

Change a toward c and then below it. Explain the segment and empty-locus cases using triangle inequality before looking at the plot. A numerical near-equality at a probe is a diagnostic subject to rounding, not a proof about every point. Record the symbolic parameter condition as part of your conclusion.

Fixed example point	Distance to (−3,0)	Distance to (3,0)	Sum
(0,4)	5	5	10
(5,0)	8	2	10
(3,16/5)	34/5	16/5	10

Understand

Write r₁=√((x+c)²+y²) and r₂=√((x−c)²+y²). These are actual nonnegative distances. If r₁+r₂=2a, subtracting their squares gives r₁²−r₂²=4cx. Factor the left side as (r₁−r₂)(r₁+r₂). Since a>0, divide by 2a to get r₁−r₂=2cx/a. Adding and subtracting the two linear equations gives

r₁=a+cx/a, and r₂=a−cx/a.

Square the first expression and compare with r₁²=(x+c)²+y². The 2cx terms cancel, leaving y²=(a²−c²)(1−x²/a²). Since b²=a²−c²>0, divide to obtain x²/a²+y²/b²=1. Thus every point of the focal locus satisfies the quadratic equation.

For the reverse implication, start with that quadratic equation. It implies ∣x∣≤a. Therefore both a+cx/a and a−cx/a are at least a−c, which is strictly positive. Using the equation to replace y² gives

(x+c)²+y²=(a+cx/a)² and (x−c)²+y²=(a−cx/a)².

Taking nonnegative square roots is now justified: the right-hand expressions before squaring were positive. Hence the actual distances equal a+cx/a and a−cx/a, and their sum is 2a. This proves the reverse direction and excludes extraneous points. Saying only “square twice and simplify” would miss the sign issue.

For a=5,c=3, b²=25−9=16, giving the stated equation. The locus is translated by replacing x,y with x−h,y−k when the center moves to (h,k); the proof uses distances from that new center.

Apply

Derive a fresh ellipse from its focal separation and distance sum, then check both directions. Identify a, c and b before writing denominators. The focal coordinate c is not generally the semiminor length b.

If a proposed equation has nonpositive b² in this horizontal form, do not call it the intended ellipse. Return to the distance definition and classify the data. When a=c>0, equality in triangle inequality restricts P to the focal segment. When a < c, there are no points. These are meaningful boundary outcomes, not numerical plotting failures.

Continue

The hyperbola uses an absolute difference of focal distances instead of their sum. That change reverses the parameter inequality and produces two branches. Keep the locus definition alongside the equation so that signs and geometric meaning remain visible.

Optional depth: when the two foci coincide, c=0, the distance-sum condition becomes twice the distance to the center equal to 2a, giving a circle. The standard nondegenerate two-focus derivation assumed c>0 for clarity, but the circular limiting case has a direct interpretation.