Ellipses: Two Foci and a Constant Sum

A fixed total distance to two points

A circle keeps one distance fixed, and a parabola makes two different kinds of distance equal. An ellipse instead fixes the sum of the distances to two points. As a moving point gets closer to one focus, it can get farther from the other, provided the total stays the same. This creates a closed oval rather than an open curve.

The inequality follows from the triangle inequality: a route from one focus to the other through P cannot be shorter than the direct segment joining the foci. If the sum equals that segment length, only points on the joining segment satisfy the condition. If it is smaller, no point does. A genuine oval therefore requires a larger sum.

The centre is the midpoint between the foci. The major axis is the longest diameter, passing through the foci; its endpoints are the vertices. The minor axis passes through the centre perpendicular to the major axis. Denote their full lengths by 2a and 2b, and the distance between the foci by 2c. Thus a and b are half-axis lengths, and c is a centre-to-focus distance, not the full focal separation.

Why the distance sum is 2a

At a vertex on the major axis, the two distances to the foci are a − c and a + c. Their sum is 2a. At an endpoint of the minor axis, symmetry makes the two focal distances equal. Each is the hypotenuse of a right triangle with legs b and c. Comparing these two special positions gives the central parameter relationship.

At the minor-axis endpoint the sum is 2√(b² + c²). Since every point has the same sum, 2√(b² + c²) = 2a. Divide by 2 and square: a² = b² + c². This proves that the foci lie inside the vertices. It also tells us how the minor axis shrinks when the foci move farther apart while a is held fixed.

Eccentricity measures focal spread

Two ellipses can have the same major-axis length but look different because their foci are separated by different amounts. Eccentricity compares focal distance with the semi-major axis. It is a ratio of lengths, so it has no unit and does not change if the entire figure is enlarged by the same scale factor.

Do not use c/b in place of c/a: the denominator is the distance to a major-axis vertex. Nor should e be confused with focal separation. If a = 5 and c = 4, e = 4/5 while the full separation is 8. They describe different quantities.

Deriving the horizontal standard equation

Choose the centre as the origin and place the foci F₁(−c, 0) and F₂(c, 0) on the x-axis. For P(x, y), write d₁ = PF₁ and d₂ = PF₂. The defining condition is d₁ + d₂ = 2a. We will isolate one distance before squaring, so that each algebraic step has a clear purpose.

The distance formula gives d₁² = (x + c)² + y² and d₂² = (x − c)² + y². Subtracting them yields d₁² − d₂² = 4cx. From d₁ = 2a − d₂, squaring gives d₁² − d₂² = 4a² − 4ad₂. Equate these expressions and divide by 4a: d₂ = a − cx/a.

Square this expression and use the distance formula again: (x − c)² + y² = (a − cx/a)². Expanding gives x² − 2cx + c² + y² = a² − 2cx + c²x²/a². Cancel −2cx, move terms and factor: x²(1 − c²/a²) + y² = a² − c². Since a² − c² = b², this becomes b²x²/a² + y² = b².

Squaring could conceal a sign condition, so verify the converse. A point satisfying the standard equation has |x| ≤ a. Substituting y² = b²(1 − x²/a²) into the focal-distance expressions gives d₁² = (a + cx/a)² and d₂² = (a − cx/a)². Because |cx/a| ≤ c < a, both brackets are strictly positive. Thus d₁ = a + cx/a and d₂ = a − cx/a, whose sum is 2a. The equation describes exactly the ellipse.

Orientation, bounds and symmetry

Each term in the standard equation is nonnegative. Therefore x²/a² ≤ 1 and y²/b² ≤ 1, giving −a ≤ x ≤ a and −b ≤ y ≤ b. The curve touches these bounds at its axis endpoints. Changing the sign of either coordinate leaves the squares unchanged, proving symmetry about both axes.

If the major axis is vertical, exchange the coordinate roles. The equation is x²/b² + y²/a² = 1, with vertices (0, ±a) and foci (0, ±c). In a normalised ellipse equation, the larger denominator identifies the major-axis direction. This is because its square root is the larger axis intercept, not simply because of the position of a symbol in a memorised template.

The ellipse’s latus rectum

As for a parabola, a latus rectum is a chord through a focus perpendicular to the main axis. An ellipse has one such chord through each focus. To find the horizontal ellipse’s chord through (c, 0), set x = c in its equation. Let the half-chord length be l, so the upper endpoint is (c, l).

Substitution gives c²/a² + l²/b² = 1. Hence l² = b²(1 − c²/a²). Since a² − c² = b², the bracket is b²/a², so l² = b⁴/a² and l = b²/a. Reflection in the major axis gives the second half of the chord. The full length is therefore 2b²/a.

Constructing ellipses from supplied data

First establish the major-axis direction, then identify which lengths describe a, b or c. Use a² = b² + c² to find the missing parameter. If two point conditions are given, the reciprocals 1/a² and 1/b² can serve as unknowns in a pair of linear equations. Afterwards check that the recovered axis lengths agree with the stated direction.

Check your understanding

These questions test the distinction between full and half lengths, the choice of orientation, and the constant-sum condition. In each calculation, label the roles of a, b and c before using their relationship.