The hyperbola

In the lesson «The ellipse» the sum of the distances to two foci stayed constant. Keep the difference constant instead and a very different curve appears: the hyperbola, two separate branches that run off to infinity along a pair of straight lines. You have met one before — the graph of y = k/x is a hyperbola turned through 45°. Hyperbolas are used in navigation too: the difference in the arrival times of signals from two radio stations fixes the difference of the distances to them, so the ship lies on a hyperbola. In exam questions almost everything about the hyperbola comes down to three numbers a, b, c — and to its asymptotes and eccentricity.

Definition: a constant difference of distances

Why must 2a be smaller than 2c? In triangle PF₁F₂ any side is longer than the difference of the other two, so ||PF₁| − |PF₂|| < |F₁F₂|. With 2a = 2c the «curve» becomes two rays of the line F₁F₂, and with 2a > 2c there are no such points at all. The absolute value matters as well: the points with |PF₁| − |PF₂| = 2a form only the branch nearer to F₂, and the points with |PF₂| − |PF₁| = 2a form the other branch.

The midpoint of F₁F₂ is the centre of the hyperbola. The curve crosses the line of the foci at two vertices; the segment between them is the real axis (also called the transverse axis), of length 2a. The perpendicular segment of length 2b through the centre is the imaginary axis (conjugate axis): the curve never meets it, but b fixes the asymptotes. The distance between the foci, 2c, is the focal distance.

Standard equations and c² = a² + b²

Put the foci on the x-axis at F₁(−c, 0) and F₂(c, 0), write the definition with the distance formula, get rid of the roots by squaring twice and denote b² = c² − a². The result is the first equation below — exactly the ellipse calculation with one sign changed. If the foci lie on the y-axis, x and y swap places.

How do you see where the foci are? In an ellipse the larger denominator shows the major axis, but in a hyperbola the sign decides: the foci lie on the axis of the variable with the plus sign. In y²/4 − x²/9 = 1 we have a² = 4 and b² = 9 — a is smaller than b, and that is perfectly allowed. More generally, mx² + ny² = 1 is a hyperbola exactly when m and n have opposite signs (mn < 0).

Asymptotes

Solve x²/a² − y²/b² = 1 for y: y = ±(b/a)·√(x² − a²). For large |x| the root √(x² − a²) is almost |x|, so the branches come closer and closer to the lines y = ±(b/a)x without ever touching them. These lines are the asymptotes (渐近线). A quick picture: draw the rectangle with sides 2a and 2b around the centre; its diagonals are the asymptotes, and its half-diagonal √(a² + b²) equals c, so the foci lie on the circle through the corners of the rectangle.

One more fact that saves time: the distance from the focus (c, 0) to the asymptote bx − ay = 0 is |bc|/√(b² + a²) = bc/c = b. So «the distance from a focus to an asymptote» is simply the imaginary semi-axis.

Eccentricity and the equilateral hyperbola

The formula links e with the asymptotes, so an exam can give one and ask for the other. If an asymptote makes the angle θ with the real axis, then b/a = tan θ and e = √(1 + tan²θ) = 1/cos θ: θ = 45° gives e = √2, θ = 60° gives e = 2. Always check first on which axis the foci lie — the same asymptotes give a different e for the other position.

The graph of y = k/x (k ≠ 0) is an equilateral hyperbola too: it is x² − y² = 2k turned through 45°, its asymptotes are the coordinate axes and again e = √2. For y = 2/x, for example, the vertices are (√2, √2) and (−√2, −√2) and the foci are (2, 2) and (−2, −2). The converse is useful as well: perpendicular asymptotes ⇔ a = b ⇔ e = √2.

How CSCA asks about this

• Asymptotes ↔ eccentricity: one is given, the other is asked; the position of the foci decides between e² = 1 + k² and e² = 1 + 1/k².
• Equation from data: foci and a vertex, a focus and an asymptote, asymptotes and a point (the λ-family).
• The definition: |PF₂| from |PF₁| with the check «at least c − a»; the triangle F₁PF₂ with a right angle at P (S = b²).
• Parameters: when is mx² + ny² = 1 or x²/(…) + y²/(…) = 1 a hyperbola, and on which axis are its foci.
• Equilateral hyperbola: a = b, asymptotes y = ±x, e = √2; «perpendicular asymptotes» says the same thing.
• Mixed with the ellipse: a hyperbola with the same foci as an ellipse — find c with the ellipse rule there (see «The ellipse») and with c² = a² + b² here.