Derive the equations of ellipses and hyperbolas from their foci using the constant sum or difference of distances

Learning goal: Derive the equations of ellipses and hyperbolas given the foci, using the fact that the sum of distances from the foci is constant for an ellipse and the difference of distances from the foci is constant for a hyperbola.

Scope note. G.GPE.3 is marked (+) in Ohio's Learning Standards for mathematics, which Ohio uses to indicate a standard "for students who are planning on taking advanced courses". It is taught here in full as part of the Geometry course, as extension content for students on that path.

Two definitions built from two points. Fix two points F₁ and F₂, the foci. An ellipse is the set of points P for which the sum PF₁ + PF₂ is a fixed constant, which we call 2a, with 2a greater than the distance between the foci. (A string of length 2a with its two ends pinned at the foci, pulled taut by a pencil, traces an ellipse.) A hyperbola is the set of points P for which the difference |PF₁ − PF₂| is a fixed constant 2a, with 2a less than the distance between the foci. The distance between the foci is written 2c, so c is the distance from the center (the midpoint of the foci) to each focus. For an ellipse a > c; for a hyperbola 0 < a < c (with a = 0 the points equidistant from the foci form the perpendicular bisector, not a hyperbola).

Deriving the ellipse. Put the foci at (−4, 0) and (4, 0), so c = 4, and let the constant sum be 10, so a = 5. For P(x, y): √((x + 4)² + y²) + √((x − 4)² + y²) = 10. Isolate one root: √((x + 4)² + y²) = 10 − √((x − 4)² + y²). Square both sides: (x + 4)² + y² = 100 − 20√((x − 4)² + y²) + (x − 4)² + y². Expand the squares and cancel: (x + 4)² − (x − 4)² = 16x, so 16x = 100 − 20√((x − 4)² + y²). Divide by 4 and isolate the root: 5√((x − 4)² + y²) = 25 − 4x. Square again: 25((x − 4)² + y²) = 625 − 200x + 16x², that is 25x² − 200x + 400 + 25y² = 625 − 200x + 16x². Collect: 9x² + 25y² = 225. Divide by 225: x²/25 + y²/9 = 1. Squaring can in principle admit extra points, so check that the steps reverse: every point of x²/25 + y²/9 = 1 has |x| ≤ 5, so 25 − 4x ≥ 5 > 0 and √((x − 4)² + y²) = (25 − 4x)/5 ≤ 9 < 10; both squared sides were therefore nonnegative, and every point of the equation has distance sum exactly 10. The denominators are a² = 25 and b² = 9, where b² = a² − c² = 25 − 16 = 9. The general result for foci (±c, 0) and sum 2a is x²/a² + y²/b² = 1 with b² = a² − c².

Reading the ellipse. Setting y = 0 gives x = ±5, the vertices, at the ends of the major axis of length 2a = 10; setting x = 0 gives y = ±3, the ends of the minor axis of length 2b = 6. The foci sit on the major axis, c = 4 from the center. A vertex confirms the constant: (5, 0) is 1 from (4, 0) and 9 from (−4, 0), sum 10. So does a minor-axis endpoint: (0, 3) is √(16 + 9) = 5 from each focus, sum 10, which is also why b² = a² − c²: a right triangle with legs b and c has hypotenuse a.

A worked model inside the lesson. Verify that (3, 2.4) is on the ellipse x²/25 + y²/9 = 1 by both descriptions. Equation: 9/25 + 5.76/9 = 0.36 + 0.64 = 1. Distances: to (4, 0), √((3 − 4)² + 2.4²) = √(1 + 5.76) = √6.76 = 2.6; to (−4, 0), √(49 + 5.76) = √54.76 = 7.4. Sum: 2.6 + 7.4 = 10. Both descriptions agree, which is what the derivation guarantees.

Deriving the hyperbola. Put the foci at (−5, 0) and (5, 0), so c = 5, and let the constant difference be 8, so a = 4. For P(x, y): √((x + 5)² + y²) − √((x − 5)² + y²) = ±8 (the sign depends on which focus is closer, which is why the definition uses the absolute value). Isolate: √((x + 5)² + y²) = ±8 + √((x − 5)² + y²). Square: (x + 5)² + y² = 64 ± 16√((x − 5)² + y²) + (x − 5)² + y². Since (x + 5)² − (x − 5)² = 20x, this gives 20x − 64 = ±16√((x − 5)² + y²), or 5x − 16 = ±4√((x − 5)² + y²). Square again (the ± disappears): 25x² − 160x + 256 = 16(x² − 10x + 25 + y²) = 16x² − 160x + 400 + 16y². Collect: 9x² − 16y² = 144. Divide by 144: x²/16 − y²/9 = 1. The ± sign covered both branches before each squaring, and the steps reverse in the same way (every point of the equation has |x| ≥ 4), so every point of the equation has distance difference exactly 8. Here a² = 16 and b² = c² − a² = 25 − 16 = 9. The general result for foci (±c, 0) and difference 2a is x²/a² − y²/b² = 1 with b² = c² − a².

Reading the hyperbola. Setting y = 0 gives x = ±4, the vertices; setting x = 0 gives −y²/9 = 1, impossible, so the curve never crosses the y-axis: it has two separate branches, one opening right and one opening left. Solving for y for large x, y ≈ ±(3/4)x, the asymptotes y = ±(b/a)x, which the branches approach but never touch. Check the constant with the vertex (4, 0): distances 9 and 1, difference 8. Check a point with y ≠ 0: (5, 2.25) gives 25/16 − 5.0625/9 = 1.5625 − 0.5625 = 1; its distances are 2.25 to (5, 0) and √(100 + 5.0625) = √105.0625 = 10.25 to (−5, 0), difference 8.

Foci on the y-axis. If the foci are (0, ±c), the same derivations with x and y exchanged give y²/a² + x²/b² = 1 for the ellipse and y²/a² − x²/b² = 1 for the hyperbola. Foci (0, ±3) with sum 10 give a = 5, b = 4 and x²/16 + y²/25 = 1, a tall ellipse whose vertices (0, ±5) are 2 and 8 from the foci. Foci (0, ±13) with difference 10 give a = 5, b = 12 and y²/25 − x²/144 = 1, whose branches open up and down. With y and x exchanged, the asymptotes of y²/a² − x²/b² = 1 are y = ±(a/b)x, here y = ±(5/12)x.

The misconception: the two derivations are the same with a sign change, so the formulas are the same. The sign in the definition changes which of a and c is larger, and that changes b². For the ellipse a > c and b² = a² − c²; for the hyperbola c > a and b² = c² − a². Mixing them up gives a negative b². The term with a² is always the positive one under the variable along which the foci lie. A second misconception is squaring only once; the two roots require two squarings, and each squaring must be done to a whole side, never term by term.

Worked ellipse derivation, second example: Foci (±3, 0), constant sum 10, so c = 3 and a = 5. Then b² = 25 − 9 = 16 and the equation is x²/25 + y²/16 = 1. Verify with (0, 4): 0 + 16/16 = 1; its distances to (3, 0) and (−3, 0) are both √(9 + 16) = 5, sum 10. Verify with (3, 3.2): 9/25 + 10.24/16 = 0.36 + 0.64 = 1; distances √(0 + 10.24) = 3.2 and √(36 + 10.24) = √46.24 = 6.8, sum 10.

Worked hyperbola derivation, second example: Foci (±13, 0), constant difference 24, so c = 13, a = 12, b² = 169 − 144 = 25, and the equation is x²/144 − y²/25 = 1. Vertices (±12, 0); asymptotes y = ±(5/12)x. Verify the vertex (12, 0): distances 25 and 1, difference 24. Verify (13, 25/12): 169/144 − (625/144)/25 = 169/144 − 25/144 = 144/144 = 1; distances 25/12 ≈ 2.083 to (13, 0) and √(676 + 625/144) = √(97344/144 + 625/144) = √(97969/144) = 313/12 ≈ 26.083 to (−13, 0), difference 288/12 = 24.

Worked from equation to foci: x²/49 + y²/24 = 1 is an ellipse with a² = 49 and b² = 24 (a² is the larger denominator, under x, so the foci are on the x-axis). c² = a² − b² = 25, c = 5: foci (±5, 0), constant sum 2a = 14. For y²/9 − x²/16 = 1, a hyperbola with a² = 9 under y: c² = a² + b² = 25, c = 5, foci (0, ±5), constant difference 2a = 6, vertices (0, ±3).

Worked check of the constraint on the constant: Foci (±4, 0) with a "constant sum" of 6 give a = 3 < c = 4 and b² = 9 − 16 = −7. No ellipse exists, because the sum of distances from any point to the two foci is at least the distance between them, 8, by the triangle inequality. A sum of 6 is impossible; the set is empty.

Derivation record: Write out the ellipse derivation for foci (±4, 0) and sum 10, and the hyperbola derivation for foci (±5, 0) and difference 8, one per page, with a one-line justification beside each step (isolate a root, square both sides, expand, collect, divide). Then make a two-column comparison: definition, relation among a, b, c, standard equation, vertices, and how you know which axis the foci lie on.

Independent performance: Choose foci (±c, 0) and a constant sum of your own with a > c, derive the ellipse equation from the definition, and verify two points by both the equation and the distance sum. Then choose foci (0, ±c) and a constant difference with 0 < a < c, derive the hyperbola, give its vertices and asymptotes, and verify one point on each branch by the distance difference. Success criteria: both squarings are applied to whole sides; b² is computed with the correct sign convention and stated; every verification shows the distance computations to two decimal places when irrational; and the write-up explains why an ellipse needs 2a greater than 2c and a hyperbola needs 2a less than 2c.