Hyperbolas
Textbook Reference
| Primary source | OpenStax Calculus Volume 2, Section 7.5: “Conic Sections” |
| Direct link | https://openstax.org/books/calculus-volume-2/pages/7-5-conic-sections |
| Textbook used in class | Stewart, Calculus, Section 10.5: “Conic Sections” (Examples 3, 4) |
Opening Scenario
A sonic boom forms when an aircraft travels faster than sound. The pressure wave on the ground has the shape of a hyperbola. Long-range navigation systems locate ships by measuring the difference in arrival times of radio signals from two stations; each measured difference corresponds to one branch of a hyperbola, and the ship sits at the intersection of two such branches. Both applications rest on the defining property: every point on a hyperbola has the same absolute difference of distances from two fixed foci.
Quick Reference
Definition. A hyperbola is the set of all points $P$ such that the absolute value of the difference of distances from $P$ to two fixed foci $F_1$ and $F_2$ equals a constant $2a$: \[ \bigl||PF_1| - |PF_2|\bigr| = 2a. \]
Standard forms with center at the origin:
| Axis | Equation | Vertices | Asymptotes |
|---|---|---|---|
| $x$-axis (opens left/right) | $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ | $(\pm a, 0)$ | $y = \pm\dfrac{b}{a}x$ |
| $y$-axis (opens up/down) | $\dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1$ | $(0, \pm a)$ | $y = \pm\dfrac{a}{b}x$ |
Key relationship (note the plus sign -- different from the ellipse!): \[ c^2 = a^2 + b^2, \quad \text{so } c = \sqrt{a^2 + b^2}. \] Since $c^2 > a^2$, we have $c > a$: the foci lie outside the vertices.
Eccentricity: $e = \dfrac{c}{a} > 1$ for every hyperbola.
Key Concepts
1. The Difference-of-Distances Definition
A hyperbola consists of two separate branches. On one branch, the point is closer to $F_1$; on the other, it is closer to $F_2$. The constant $2a$ is the magnitude of the difference.
This is the key contrast with an ellipse: an ellipse uses a sum ($|PF_1| + |PF_2| = 2a$) and forms a closed curve; a hyperbola uses a difference and forms two open branches.
2. Reading the Parameters From the Equation
The positive term in the standard form tells you which axis the hyperbola opens along. If $x^2$ is positive, the hyperbola opens left and right; if $y^2$ is positive, it opens up and down.
Example 1. Find the vertices, foci, and asymptotes of $x^2/4 - y^2/9 = 1$. (This is Stewart 10.5, Example 3.)
Goal. The equation is already in standard form. Read off $a$, $b$, $c$.
The positive term is $x^2/4$, so the hyperbola opens left and right. Here $a^2 = 4$ and $b^2 = 9$, so \[ a = 2, \quad b = 3, \quad c = \sqrt{4 + 9} = \sqrt{13}. \]
Boxed answer:
- Vertices: $(\pm 2,\, 0)$.
- Foci: $(\pm\sqrt{13},\, 0)$.
- Asymptotes: $y = \pm\dfrac{3}{2}x$.
Recap. For a horizontal hyperbola, the asymptote slope is $b/a$ (rise $b$, run $a$), not $a/b$. The asymptotes do not intersect the hyperbola; they are the diagonal lines the branches approach as $x \to \pm\infty$.
3. Finding the Equation From the Geometric Data
Example 2. Find the equation of the hyperbola with foci $(\pm 5, 0)$ and vertices $(\pm 3, 0)$. (This is Stewart 10.5, Example 4.)
Goal. The foci and vertices both lie on the $x$-axis, so the hyperbola opens left and right. Use $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$.
From the vertices: $a = 3$, so $a^2 = 9$. From the foci: $c = 5$, so $b^2 = c^2 - a^2 = 25 - 9 = 16$.
\[ \frac{x^2}{9} - \frac{y^2}{16} = 1. \]
Boxed answer: $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$.
Asymptotes: $y = \pm\dfrac{4}{3}x$.
Recap. With a hyperbola you compute $b^2 = c^2 - a^2$ (because $c > a$, this is positive). With an ellipse you compute $b^2 = a^2 - c^2$ (because $c < a$, this is positive). Getting the order of subtraction wrong is the single most common error when switching between the two conic types.
4. The Asymptotes and How to Draw Them
The asymptotes are not part of the hyperbola but they are essential for sketching it. The reliable method:
- Draw the central rectangle of width $2a$ (horizontal) and height $2b$ (vertical), centered at the origin.
- The asymptotes are the diagonals of that rectangle.
- Sketch the two branches, one opening through each pair of vertices, curving toward the asymptotes.
For $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$: the rectangle has corners $(\pm a, \pm b)$, and the asymptotes pass through those corners.
confusing $c^2 = a^2 + b^2$ for a hyperbola with $c^2 = a^2 - b^2$ for an ellipse. The formulas look similar but differ in sign. The mnemonic is: for an ellipse the foci are trapped inside the oval ($c < a$), so the formula subtracts; for a hyperbola the foci are outside the curve ($c > a$), so the formula adds. If you always know which is which, the sign never trips you up.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Using $c^2 = a^2 - b^2$ for a hyperbola | $c = \sqrt{9 - 16}$ -- imaginary | Hyperbola: $c^2 = a^2 + b^2$ |
| Swapping the asymptote slopes | writing $y = \pm(a/b)x$ for a horizontal hyperbola | Horizontal hyperbola: $y = \pm(b/a)x$ |
| Calling the $b^2$ term the “semi-major axis” | $a = 3, b = 4$, but claiming $b$ is the major semi-axis | $a$ is always the semi-transverse axis (vertex distance); $b$ sets the asymptote slope |
| Forgetting $b^2 = c^2 - a^2$ when finding the equation | computing $a^2 - c^2 < 0$ | $b^2 = c^2 - a^2 > 0$ because $c > a$ for a hyperbola |
| Trying to subtract first in the equation and losing the standard form | writing $-y^2/b^2 + x^2/a^2 = 1$ from the $y$-axis form | The $y$-axis form is $y^2/a^2 - x^2/b^2 = 1$, not the negated version |
Common Misconceptions
for a hyperbola $c^2 = a^2 - b^2$ (using the ellipse formula).
This is the concept-image-conflicts-definition error caused by confusing the two conic formulas. For a hyperbola the foci lie outside the curve, so $c > a$, which forces $c^2 = a^2 + b^2$. A counterexample: for $\frac{x^2}{9} - \frac{y^2}{16} = 1$, the correct focal distance is $c = \sqrt{9 + 16} = 5$; using $c^2 = 9 - 16 = -7$ produces an imaginary value, showing immediately that the formula is wrong.
the asymptote slope of a horizontal hyperbola $x^2/a^2 - y^2/b^2 = 1$ is $\pm a/b$.
This is the input-output-confusion error in reading the asymptote formula. The asymptotes of the horizontal hyperbola are $y = \pm (b/a)x$: the $b$ is in the numerator and $a$ is in the denominator. Swapping them gives the asymptotes of the vertical form instead. The mnemonic is that the asymptote slope uses $b$ over $a$ when the positive term is $x^2/a^2$.
Leveled Practice
Level 1 -- Direct Reading
Problem 1. Write $4y^2 - 9x^2 = 36$ in standard form and identify the vertices, foci, and asymptotes.
Show answer
Divide by $36$: $\dfrac{y^2}{9} - \dfrac{x^2}{4} = 1$.
The positive term is under $y^2$, so the hyperbola opens up and down. $a^2 = 9$, $b^2 = 4$, $c = \sqrt{9 + 4} = \sqrt{13}$.
- Vertices: $(0,\, \pm 3)$.
- Foci: $(0,\, \pm\sqrt{13})$.
- Asymptotes: $y = \pm\dfrac{3}{2}x$ (slope $= a/b = 3/2$ for a vertical hyperbola).
Problem 2. A hyperbola has vertices $(\pm 2, 0)$ and asymptotes $y = \pm\dfrac{5}{2}x$. Find its equation and its foci.
Show answer
Horizontal hyperbola, so $a = 2$ and the slope is $b/a = 5/2$, giving $b = 5$.
$c = \sqrt{4 + 25} = \sqrt{29}$.
Boxed answer: $\dfrac{x^2}{4} - \dfrac{y^2}{25} = 1$, foci at $(\pm\sqrt{29},\, 0)$.
Level 2 -- Multi-Step
Problem 3. Find the equation of the hyperbola that passes through $(3, 2)$ and has foci $(\pm 4, 0)$.
Show answer
The foci are on the $x$-axis with $c = 4$, so the form is $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ with $b^2 = 16 - a^2$.
Substituting the point $(3, 2)$: \[ \frac{9}{a^2} - \frac{4}{16 - a^2} = 1. \] Multiply through by $a^2(16 - a^2)$: \[ 9(16 - a^2) - 4a^2 = a^2(16 - a^2). \] \[ 144 - 9a^2 - 4a^2 = 16a^2 - a^4. \] \[ a^4 - 29a^2 + 144 = 0. \] Let $u = a^2$: $u = (29 \pm \sqrt{841 - 576})/2 = (29 \pm \sqrt{265})/2$.
Both roots are positive; we need $0 < a^2 < 16$ (so $b^2 > 0$). $\sqrt{265} \approx 16.3$, so $(29 - 16.3)/2 \approx 6.35$ and $(29 + 16.3)/2 \approx 22.65$. Take $a^2 \approx 6.35$ (the other gives $b^2 < 0$). The exact value is $a^2 = (29 - \sqrt{265})/2$.
(This problem shows that general conics do not always have clean parameters.)
Mastery Checklist
Mental Model
Compare the three conics by their eccentricity $e = c/a$:
| Conic | Eccentricity | Foci relative to vertices |
|---|---|---|
| Circle | $e = 0$ | foci coincide at center |
| Ellipse | $0 < e < 1$ | foci inside the curve |
| Parabola | $e = 1$ | one focus, directrix at infinity |
| Hyperbola | $e > 1$ | foci outside the curve |
For an ellipse, $c < a$, so the Pythagorean relationship subtracts: $c^2 = a^2 - b^2$. For a hyperbola, $c > a$, so the relationship adds: $c^2 = a^2 + b^2$. Remembering which inequality holds for $c$ versus $a$ makes the sign automatic.
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