Hold the difference of distances fixed, and two branches fly apart.
An ellipse came from a constant sum of distances to two pins. Change one word — sum to difference — and the curve breaks in two. Keep the absolute difference | |PF₁| − |PF₂| | pinned at a fixed length 2a, and the path you trace is a hyperbola: two open branches arcing away from each other, each hugging a pair of straight asymptotes. Its equation x²/a² − y²/b² = 1 wears one plus and one minus, and — watch closely — its right triangle flips the ellipse's: now c² = a² + b², so c is the largest and the foci sit outside the vertices.
Pin two points, the foci F₁ and F₂. Now demand of a moving point P not that the sum of its distances stay fixed (that was the ellipse, 30.5), but that the difference stay fixed:
| |PF₁| − |PF₂| | = 2a (a constant).
The vertical bars around the difference matter. On the right branch, P is closer to F₂, so |PF₁| − |PF₂| = +2a. On the left branch, P is closer to F₁, so the bare difference is −2a. The absolute value folds both signs into one rule — and that is exactly why the curve has two branches, one for each sign.
A hyperbola is the set of points whose distances to two fixed foci have a constant absolute difference. For the difference to be a real curve we need 2a < |F₁F₂| — the foci must be farther apart than the gap they prescribe. (If 2a = |F₁F₂| you get the two rays outward; if 2a > |F₁F₂| nothing satisfies it.)
Set the curve upright: centre at the origin, foci on the x-axis at (±c, 0). Writing | |PF₁| − |PF₂| | = 2a in coordinates, squaring twice to clear the radicals, and abbreviating b² = c² − a², the dust settles into a clean equation:
x2a2 − y2b2 = 1
The signature is the minus sign. An ellipse is plus–plus; a hyperbola is plus–minus. The variable with the plus tells you which way it opens: x²/a² − y²/b² = 1 opens left–right (the x-term is positive), crossing the x-axis at the vertices (±a, 0). Flip the signs to y²/a² − x²/b² = 1 and it opens up–down instead, crossing the y-axis at (0, ±a).
Here a is not the bigger number, the way it was for the ellipse — a is simply the semi-distance to the vertices (it sits under the positive term). For the horizontal hyperbola the curve never visits |x| < a at all: there is an empty corridor between the branches.
For the ellipse the right triangle of half-axes gave a² = b² + c², with a the hypotenuse (the largest). The hyperbola turns this inside out. Setting b² = c² − a² rearranges to the Pythagorean-looking fact
c2 = a2 + b2
so now c is the hypotenuse — the largest of the three. The foci (±c, 0) therefore lie outside the vertices (±a, 0): the curve opens, and the foci are tucked inside each branch's bend. Picture the right triangle with legs a and b standing on a vertex: its hypotenuse reaches exactly out to a focus.
For x²/9 − y²/16 = 1 we read a² = 9, b² = 16, so a = 3, b = 4 and c = √(9 + 16) = √25 = 5. Vertices at (±3, 0), foci at (±5, 0) — the foci sit beyond the vertices, as they must. Compare the ellipse x²/25 + y²/16 = 1, where c = √(25 − 16) = 3 lands the foci inside. Same three letters, opposite triangle.
Solve x²/a² − y²/b² = 1 for y and you get y = ±(b/a)√(x² − a²). For large |x| the “− a²” barely matters, and √(x² − a²) ≈ |x|, so the branches creep ever closer to the two straight lines
y = ±ba x
These are the asymptotes: guide lines the curve approaches but never reaches. They are the diagonals of the central rectangle 2a wide and 2b tall, and they fix the curve's flare — a steeper b/a means branches that open more sharply. Drawn slate and dashed, they are scaffolding, not part of the hyperbola.
The asymptotes belong to both forms: y = ±(b/a)x for the horizontal hyperbola, and y = ±(a/b)x for the vertical one. A quick way to get them: replace the 1 on the right with 0, giving x²/a² − y²/b² = 0, which factors into the two lines y = ±(b/a)x.
Just like the ellipse, the hyperbola measures its shape with
e = ca
but here c > a, so e > 1 — always. (The ellipse had 0 < e < 1; the parabola, next lesson, sits exactly at e = 1.) Since the asymptote slope is b/a = √(c² − a²)/a = √(e² − 1), a larger e means steeper asymptotes and wider-flaring branches. As e → 1⁺ the branches pinch toward two nearly straight rays; as e grows large they swing open toward a right angle.
One number e = c/a names the whole conic family: e < 1 ellipse, e = 1 parabola, e > 1 hyperbola. The hyperbola lives entirely in the e > 1 territory — it is the only conic with two branches and asymptotes.
The two central conics are mirror cousins. Loop a string and you get one bounded oval; demand a difference and you get two unbounded branches. The single sign change in the equation drags everything else along with it — which axis the foci pile up on, which way the a–b–c triangle tips, and whether the eccentricity sits below or above 1.
| Ellipse (30.5) | Hyperbola (30.6) | |
|---|---|---|
| defining rule | |PF₁| + |PF₂| = 2a (sum) | | |PF₁| − |PF₂| | = 2a (difference) |
| equation | x²/a² + y²/b² = 1 | x²/a² − y²/b² = 1 |
| triangle | a² = b² + c² (a largest) | c² = a² + b² (c largest) |
| foci vs vertices | foci inside (c < a) | foci outside (c > a) |
| eccentricity | e = c/a, 0 < e < 1 | e = c/a, e > 1 |
| asymptotes | none (bounded) | y = ±(b/a)x |
Read the table left to right and the single flip — plus to minus, b² + c² to a² + b² — explains every other difference. Memorize the triangles together so you never swap them: a² = b² + c² for the ellipse, c² = a² + b² for the hyperbola.
| idea | the fact |
|---|---|
| defining property | | |PF₁| − |PF₂| | = 2a (constant absolute difference) |
| standard equation | x²/a² − y²/b² = 1 (opens left–right); y²/a² − x²/b² = 1 (up–down) |
| vertices & foci | vertices (±a, 0); foci (±c, 0) with c² = a² + b² |
| asymptotes | y = ±(b/a)x — approached, never reached; not part of the curve |
| eccentricity | e = c/a > 1; bigger e ⇒ steeper asymptotes, wider flare |
| vs the ellipse | ellipse: sum, a² = b² + c², e < 1, bounded — the exact opposite |
For x²/16 − y²/9 = 1, find a, b, c, the vertices, the foci, the asymptotes, and the eccentricity.
a² = 16 ⇒ a = 4; b² = 9 ⇒ b = 3. Then c = √(16 + 9) = √25 = 5. Vertices (±4, 0); foci (±5, 0). Asymptotes y = ±(b/a)x = ±(3/4)x. Eccentricity e = c/a = 5/4 = 1.25 (> 1, good).
A hyperbola has vertices at (±2, 0) and foci at (±√13, 0). Write its standard equation.
a = 2 (so a² = 4); c = √13 (so c² = 13). From c² = a² + b², b² = 13 − 4 = 9. It opens left–right (foci on the x-axis), so x²/4 − y²/9 = 1.
Which is the asymptote slope of x²/9 − y²/4 = 1, and which of the ellipse x²/9 + y²/4 = 1? Explain the difference.
The hyperbola has asymptotes y = ±(b/a)x = ±(2/3)x. The ellipse x²/9 + y²/4 = 1 has no asymptotes — it is a bounded oval (|x| ≤ 3, |y| ≤ 2). Asymptotes are a feature only of the unbounded hyperbola, whose branches run off to infinity.
A hyperbola has asymptotes y = ±2x and passes through (2, 0). Find its equation.
It crosses the x-axis at (2, 0), so it opens left–right with a vertex there: a = 2 (a² = 4). The asymptote slope is b/a = 2, so b = 2a = 4 (b² = 16). Equation: x²/4 − y²/16 = 1. (Check: c = √(4 + 16) = √20 = 2√5, e = √5 ≈ 2.24 > 1.)
A point P on a hyperbola with foci F₁(−5, 0), F₂(5, 0) satisfies |PF₁| = 9 and lies on the curve with a = 3. Find |PF₂|, and say which branch P is on.
The absolute difference is 2a = 6, so | |PF₁| − |PF₂| | = 6. With |PF₁| = 9 that gives |PF₂| = 9 − 6 = 3 or |PF₂| = 9 + 6 = 15. The minimum distance to F₂ is c − a = 5 − 3 = 2, and on the right branch (nearer F₂) the closest point is the vertex with |PF₂| = c − a = 2. Since |PF₁| > |PF₂| here, P is on the right branch with |PF₂| = 3 (the value ≥ c − a = 2 and consistent with the +2a sign).
Without swapping the formulas: an ellipse and a hyperbola both have a = 5 and c = 13. Find b for each. What goes wrong if you mix the triangles?
Ellipse needs a largest (a > c), but here a = 5 < c = 13 — impossible for an ellipse (e = c/a = 2.6 would exceed 1). Hyperbola: c² = a² + b² ⇒ b² = 169 − 25 = 144, b = 12; e = 13/5 = 2.6 > 1, valid. The lesson: with c the largest you must be on the hyperbola triangle c² = a² + b²; using the ellipse's a² = b² + c² here gives b² = 25 − 169 < 0, a red flag that you swapped them.
Six questions to lock it in. Tap the answer you think is right.
This lesson develops the hyperbola as a locus (the constant absolute difference of focal distances), derives the standard form x²/a² − y²/b² = 1, and establishes c² = a² + b², the asymptotes y = ±(b/a)x, and the eccentricity e = c/a > 1. It addresses HSG-GPE.A.3 (derive the equation of a hyperbola from the definition as the difference of distances to the foci) and reinforces HSA-REI reasoning (solving the defining equation and relating a, b, c). The deliberate side-by-side with the ellipse (30.5) targets the most common student error — swapping a² = b² + c² with c² = a² + b². Every figure is computed from a, b, c by the page's geometry engine, so the picture and the numbers always agree; encourage learners to drive the sliders and confirm the difference of distances really holds constant.