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Derive standard equation of an ellipse

Short answer

Sum of distances to foci (±c,0) is 2a. Squaring √((x+c)²+y²)+√((x-c)²+y²)=2a gives x²/a² + y²/b²=1 with b²=a²-c².

Derive the equation of an ellipse in standard form.

An ellipse is the locus of a point P that moves so that the sum of its distances from two fixed points (foci) is constant and greater than the distance between the foci. (Equivalently, the distance from a focus is e times the distance from a directrix, with 0 < e < 1.)

Derivation

Take the foci F₁(−c, 0) and F₂(c, 0) and let the constant sum be 2a, with a > c > 0. For P(x, y) on the locus:

PF₁ + PF₂ = 2a

√((x+c)² + y²) + √((x−c)² + y²) = 2a

Move the second root across and square:

(x+c)² + y² = 4a² − 4a√((x−c)² + y²) + (x−c)² + y²

4cx = 4a² − 4a√((x−c)² + y²)

a√((x−c)² + y²) = a² − cx

Square again:

a²x² − 2a²cx + a²c² + a²y² = a⁴ − 2a²cx + c²x²

(a² − c²)x² + a²y² = a²(a² − c²)

Since a > c, put b² = a² − c² (b > 0):

b²x² + a²y² = a²b²

Divide by a²b²:

x²/a² + y²/b² = 1

Properties

  • Centre (0, 0); vertices (±a, 0) and (0, ±b).
  • Major axis 2a (along the x-axis), minor axis 2b.
  • Eccentricity e = c/a = √(1 − b²/a²) < 1; foci (±ae, 0).
  • Directrices x = ±a/e.
  • Latus rectum = 2b²/a.
        y
        |  B(0,b)
     ___|___
   /    |    \
 -A-F1--O--F2-A'--x
   \____|____/
        | B'(0,-b)

✓ Answer checked by Claude Opus 5.5 (high effort)

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