Hyperbolas: Two Foci and a Constant Difference

A constant difference instead of a sum

An ellipse keeps the sum of two focal distances fixed. A hyperbola keeps their difference fixed, always subtracting the nearer distance from the farther one. A point can satisfy this condition on either side of the foci, producing two separate branches. The absolute value is essential: it describes both branches with a single condition.

For horizontal foci F₁(−c, 0) and F₂(c, 0), a point on the right branch is farther from F₁, so PF₁ − PF₂ = 2a. On the left branch the order reverses: PF₂ − PF₁ = 2a. The two equations describe the same positive difference. Using the first equation everywhere would incorrectly discard the left branch.

The centre is the midpoint of the foci. The line through the foci is the transverse axis. The perpendicular line through the centre is the conjugate axis. The two points where the curve meets the transverse axis are the vertices. For horizontal orientation these are (−a, 0) and (a, 0). Their distances from the foci show that the fixed difference is (c + a) − (c − a) = 2a.

Parameter relationship and eccentricity

The foci lie beyond the vertices, so c > a. The parameter b records the positive difference between c² and a². A hyperbola does not have a smaller and larger half-axis in the ellipse sense: a names the half transverse length and b the half conjugate length, regardless of which is numerically larger.

The constant difference must be smaller than focal separation by the reverse triangle inequality. A zero difference produces the perpendicular bisector of the focal segment, while the limiting difference equal to focal separation produces rays along the focal line. Neither case is the ordinary two-branch curve studied here.

Deriving the horizontal standard equation

Place the centre at the origin with F₁(−c, 0) and F₂(c, 0), where c > a > 0. Begin on the right branch and write d₁ − d₂ = 2a. As with the ellipse, isolating one distance before squaring makes the calculation controlled. Afterwards we will verify both branches, including the signs hidden by squared expressions.

The squared-distance difference is d₁² − d₂² = (x + c)² − (x − c)² = 4cx. Since d₁ = 2a + d₂, squaring gives d₁² − d₂² = 4a² + 4ad₂. Equating and dividing by 4a yields d₂ = cx/a − a.

Now square and substitute d₂² = (x − c)² + y²: x² − 2cx + c² + y² = c²x²/a² − 2cx + a². Cancel the common −2cx terms and rearrange to get x²(c²/a² − 1) − y² = c² − a². Replace c² − a² by b², then divide by b².

On the left branch, the starting condition is d₂ − d₁ = 2a. Isolating and squaring produces the same final equation. Therefore the derived equation is satisfied by both branches. To establish that it introduces no extra points, we must also start from the equation and recover the absolute distance difference.

The converse needs both signs

From x²/a² − y²/b² = 1 we get y² = b²(x²/a² − 1). Substitution into the squared focal distances, using b² = c² − a², gives d₁² = (cx/a + a)² and d₂² = (cx/a − a)². Taking square roots produces absolute values, not automatically the original brackets.

The equation implies x²/a² = 1 + y²/b² ≥ 1, so x ≥ a or x ≤ −a. On the right branch x ≥ a, so cx/a ≥ c > a. Both brackets are positive, giving d₁ = cx/a + a and d₂ = cx/a − a. Their difference is 2a. This includes the right vertex x = a; the required inequality is not strictly x > a.

On the left branch x ≤ −a, so cx/a ≤ −c < −a. Both brackets are negative, giving d₁ = −cx/a − a and d₂ = −cx/a + a. Now d₂ − d₁ = 2a. Hence every equation point satisfies |d₁ − d₂| = 2a, including the left vertex. This completes the converse for the whole curve.

Orientation, symmetry and allowed regions

For a horizontal hyperbola no point lies in the strip −a < x < a. Setting x = 0 would give −y²/b² = 1, impossible for real y, so it has no real intercept on the conjugate axis. Setting y = 0 gives x = ±a, the vertices. Squared coordinates establish symmetry about both axes.

For a vertical hyperbola exchange x and y to obtain y²/a² − x²/b² = 1. Its vertices and foci are on the y-axis, and its allowed regions are y ≥ a or y ≤ −a. Normalise to right side 1 before identifying the positive term. Unlike an ellipse, the larger denominator does not identify the main direction of a hyperbola: the sign does.

When a = b, the hyperbola is called equilateral. The parameter relationship then gives c = a√2 and e = √2. Equal parameters do not make the curve a circle: the subtraction sign still produces two open branches. We keep these equations centred at the origin with coordinate axes as their transverse and conjugate axes.

Deriving the hyperbola’s latus rectum

The latus rectum is again a focal chord perpendicular to the main axis, now the transverse axis. For a horizontal hyperbola, use the focus (c, 0) and set x = c. Both endpoints lie on the right branch; the corresponding chord through (−c, 0) lies on the left branch.

At x = c, the equation gives c²/a² − y²/b² = 1. Therefore y²/b² = (c² − a²)/a² = b²/a², so y² = b⁴/a² and y = ±b²/a. The difference between the endpoint y-coordinates is 2b²/a. It matches the ellipse’s formula, although its derivation uses c² − a² instead of a² − c².

Building a hyperbola from constraints

Geometric data determines the axis direction and parameters. Check that c > a so b² is positive. A focal distance and latus rectum length can lead to a quadratic equation for a; only a positive length is admissible. A point condition can also determine the remaining denominator, but the orientation may have to be tested.

Check your understanding

First identify the positive term after normalising. Then use addition to obtain c² and check that e > 1. The final questions test the geometry behind the equation rather than calculation alone.