Hyperbola — standard forms, asymptotes, foci, eccentricity: Worked examples

Where $e = 1 + (b / a)^{2}$ comes from

Before the examples, let us earn the one formula that every angle/limit problem leans on. Look at the picture: the central box has half-width $a$ and half-height $b$ ; the focus sits at distance $c$ from the centre.

Figure — Hyperbola — standard forms, asymptotes, foci, eccentricity

Why this is the useful form: the asymptote slope is $b / a$ . So the tilt of the box diagonal alone fixes $e$ — you never need $a$ and $b$ separately (Example 3 and Example 6 both cash this in). Geometrically, $c$ is the hypotenuse of the right triangle with legs $a$ (along the $x$ -axis) and $b$ (up to the box corner); the amber line in the figure is exactly that hypotenuse.

The scenario matrix

Every hyperbola question is one of these cells. The last column names the example that covers it.

#	Case class	What makes it different	Covered by
A	Transverse axis on x (horizontal)	$+$ sign on $x^{2}$ ; foci $(\pm c, 0)$ ; slope $b / a$	Ex 1
B	Transverse axis on y (vertical)	$+$ sign on $y^{2}$ ; foci $(0, \pm c)$ ; slope $a / b$	Ex 2
C	Build equation from foci + $2 a$	reverse direction: data → equation	Ex 2
D	Info given as an angle (asymptote tilt)	need $tan$ , then $e = 1 + (b / a)^{2}$	Ex 3
E	Degenerate input (RHS $= 0$ )	curve collapses to the two asymptotes	Ex 4
F1	Limiting behaviour $b \to 0^{+}$	branches pinch onto the $x$ -axis, $e \to 1^{+}$	Ex 5
F2	Limiting behaviour $b \to \infty$	branches fan toward vertical, $e \to \infty$	Ex 6
G	Rectangular case $a = b$	perpendicular asymptotes, $e = 2$	Ex 7
H	Word problem (LORAN / navigation)	constant time-difference → constant distance-difference	Ex 8
I	Exam twist — numbers hidden, must solve for $a, b$	given $e$ and a focus, find equation	Ex 9

Recall Quick self-test before you start

Which sign relation is a hyperbola's? ::: $c^{2} = a^{2} + b^{2}$ (plus — foci sit outside the vertices) Slope of asymptote when transverse axis is horizontal ::: $\pm b / a$ Slope of asymptote when transverse axis is vertical ::: $\pm a / b$

Worked Examples

Worked example Example 1 — Case A: horizontal transverse axis, full read-off

Analyse $\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1$ : find vertices, foci, $e$ , asymptotes.

Forecast: The $+$ sign is on $x^{2}$ , so guess the branches open left–right and the foci sit on the $x$ -axis. Will $e$ be closer to $1$ or large? Since $b (= 4) > a (= 3)$ , the box is tall, so the branches should be wide — guess $e$ noticeably bigger than $1$ .

Read $a^{2} = 9 \Rightarrow a = 3$ . Why this step? The term carrying the $+$ sign names the transverse (foci) direction; its denominator is $a^{2}$ . So vertices are $(\pm 3, 0)$ .
Read $b^{2} = 16 \Rightarrow b = 4$ . Why? The other denominator is $b^{2}$ ; it fixes the box height, not any vertex.
$c^{2} = a^{2} + b^{2} = 9 + 16 = 25 \Rightarrow c = 5$ . Why? Hyperbola uses plus — foci lie beyond the vertices. Foci $(\pm 5, 0)$ .
$e = c / a = 5/3 \approx 1.667$ . Why? $e = c / a$ by definition, and $> 1$ confirms it is a hyperbola.
Asymptotes $y = \pm \frac{b}{a} x = \pm \frac{4}{3} x$ . Why? Horizontal transverse ⇒ slope is (box height)/(box width) $= b / a$ .

The figure at the top of the page (blueprint plot of the amber box with corners $(\pm 3, \pm 4)$ , cyan curve, and dashed asymptote diagonals) is this hyperbola — reuse it here.

Verify: $c^{2}$ should equal $a^{2} + b^{2}$ : $5^{2} = 25 = 9 + 16$ . ✓ And $e = 5/3 > 1$ . ✓ Corner of the box $(3, 4)$ has slope $4/3$ — matches the asymptote. ✓

Worked example Example 2 — Cases B + C: vertical axis, built from data

The foci are $(0, \pm 10)$ and the constant distance-difference is $2 a = 16$ . Write the equation, then give $e$ and the asymptotes.

Forecast: Foci are on the $y$ -axis, so guess the $+$ sign lands on $y^{2}$ and branches open up–down. Since $2 a = 16$ is close to $2 c = 20$ , the branches should be fairly narrow ( $e$ near $1$ ).

Foci on $y$ -axis ⇒ vertical form $\frac{y ^{2}}{a ^{2}} - \frac{x ^{2}}{b ^{2}} = 1$ . Why this step? The transverse axis always passes through the foci; here that axis is vertical, so $y^{2}$ gets the $+$ .
$c = 10$ (focus distance) and $a = \frac{16}{2} = 8$ . Why? $2 a$ is the given difference, so $a = 8$ ; $c$ is read off the focus.
$b^{2} = c^{2} - a^{2} = 100 - 64 = 36$ . Why? Rearranged $c^{2} = a^{2} + b^{2}$ ; $b$ is whatever the plus-rule leaves over.
Equation: $\frac{y ^{2}}{64} - \frac{x ^{2}}{36} = 1$ . Why? Substitute $a^{2} = 64, b^{2} = 36$ .
$e = c / a = 10/8 = 1.25$ . Asymptotes $y = \pm \frac{a}{b} x = \pm \frac{8}{6} x = \pm \frac{4}{3} x$ . Why does the slope flip to $a / b$ ? In the vertical form the box is $y = \pm a$ , $x = \pm b$ , so the diagonal rise is $a$ over run $b$ .

Look at the figure: the branches open up and down, the foci are the amber dots on the $y$ -axis, and the up–down box makes the asymptotes tilt more steeply than a horizontal case would.

Verify: $c^{2} = a^{2} + b^{2}$ : $100 = 64 + 36$ . ✓ $e = 1.25 > 1$ . ✓ Forecast said "narrow, $e$ near $1$ " and $1.25$ is indeed close to $1$ . ✓

Worked example Example 3 — Case D: eccentricity from an angle

A hyperbola with horizontal transverse axis has asymptotes making $\pm 3 0^{\circ}$ with the $x$ -axis. Find $e$ without knowing $a$ or $b$ separately.

Forecast: Shallow asymptotes ( $3 0^{\circ}$ ) mean narrow branches, so $e$ should be only slightly above $1$ .

Slope $= tan 3 0^{\circ} = \frac{1}{3}$ . Why $tan$ ? Slope is "rise over run" and on a right triangle with the angle at the origin, $tan (angle) = \frac{opposite (rise)}{adjacent (run)}$ — exactly the slope. So $tan 3 0^{\circ}$ is the asymptote slope.
Set that equal to $b / a$ : $\frac{b}{a} = \frac{1}{3}$ . Why? For a horizontal hyperbola the asymptote slope is $b / a$ .
$e = 1 + (\frac{b}{a})^{2} = 1 + \frac{1}{3} = \frac{4}{3} = \frac{2}{3} \approx 1.155$ . Why this formula? The section above derived $e = 1 + b^{2} / a^{2}$ from the plus-rule; it uses only the ratio $b / a$ , which is all the angle gave us — we never needed $a, b$ alone.

Look at the figure: the two dashed asymptotes make a shallow $3 0^{\circ}$ wedge with the $x$ -axis, and the amber angle arc marks $tan 3 0^{\circ}$ as the slope.

Verify: $(\frac{1}{3})^{2} = \frac{1}{3}$ ; $1 + \frac{1}{3} = \frac{4}{3}$ ; $4/3 \approx 1.1547 > 1$ . ✓ Small angle ⇒ $e$ just above $1$ , as forecast. ✓

Worked example Example 4 — Case E: degenerate hyperbola (RHS = 0)

Describe the full graph of $\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 0$ (note the $0$ , not $1$ ).

Forecast: With no " $1$ " to keep the curve away from the centre, guess the "hyperbola" shrinks onto its own asymptotes — a pair of crossing straight lines.

Factor as a difference of squares: $(\frac{x}{3} - \frac{y}{4}) (\frac{x}{3} + \frac{y}{4}) = 0$ . Why this step? Anything of the form $P^{2} - Q^{2}$ factors into $(P - Q) (P + Q)$ ; a product is zero only when a factor is zero.
So $\frac{x}{3} = \frac{y}{4}$ or $\frac{x}{3} = - \frac{y}{4}$ , i.e. $y = \pm \frac{4}{3} x$ . Why? Setting each factor to zero gives the two lines.
These are exactly the asymptotes of Example 1. Why does this happen? Replacing the $1$ by $0$ is the same "drop the $1$ " limit that the parent used to find asymptotes — here it is exact, so the graph is the asymptote pair.

Look at the figure: the two cyan lines cross at the origin; there is no gap, no branch — a fully degenerate conic.

Verify: Point $(3, 4)$ : $\frac{9}{9} - \frac{16}{16} = 1 - 1 = 0$ . ✓ Point $(3, - 4)$ : also $0$ . ✓ Both lie on $y = \pm \frac{4}{3} x$ . See also Asymptotes and Limits at Infinity for why " $1 \to 0$ " is the natural limit.

Worked example Example 5 — Case F1: limiting behaviour

$b \to 0^{+}$ (branches pinch) Fix $a = 2$ . Watch $\frac{x ^{2}}{4} - \frac{y ^{2}}{b ^{2}} = 1$ as $b \to 0^{+}$ . What happens to the branches and to $e$ ?

Forecast: Tiny $b$ means a very short box; guess the branches squeeze down toward the $x$ -axis and $e \to 1^{+}$ (nearly straight rays).

Asymptote slope $= b / a = b /2 \to 0$ . Why this step? As $b \to 0$ the box height vanishes, so its diagonals flatten toward the horizontal.
$e = 1 + b^{2} / a^{2} = 1 + b^{2} /4 \to 1 = 1^{+}$ . Why? With $b^{2} \to 0$ the term under the root $\to 1$ ; $e$ approaches $1$ from above but never reaches it.
The curve $y^{2} = b^{2} (\frac{x ^{2}}{4} - 1) \to 0$ for every fixed $x$ , so both branches collapse onto the two rays $x \geq 2$ and $x \leq - 2$ of the $x$ -axis. Why? Multiplying a bounded bracket by $b^{2} \to 0$ forces $y \to 0$ .

Look at the figure: as $b$ shrinks the cyan branches flatten toward the $x$ -axis and the box collapses to a thin sliver.

Verify: At $b = 0.1$ : slope $= 0.05$ , $e = 1 + 0.0025 \approx 1.00125$ — both tiny excesses above their limits, confirming the $1^{+}$ approach. ✓

Worked example Example 6 — Case F2: limiting behaviour

$b \to \infty$ (branches fan out) Fix $a = 2$ again. Now watch $\frac{x ^{2}}{4} - \frac{y ^{2}}{b ^{2}} = 1$ as $b \to \infty$ . What happens to the branches and to $e$ ?

Forecast: A very tall box should make the asymptotes almost vertical and $e$ blow up — the opposite of Example 5.

Asymptote slope $= b / a = b /2 \to \infty$ . Why this step? A huge box height makes the diagonals nearly vertical (steep slope).
$e = 1 + b^{2} / a^{2} = 1 + b^{2} /4 \to \infty$ . Why? With $b^{2} \to \infty$ the term under the root grows without bound, so $e$ grows without bound.
Near the vertices $(\pm 2, 0)$ the curve turns sharply and then rises almost vertically alongside the steep asymptotes. Why? For large $b$ , $y = \pm \frac{b}{2} x 1 - 4/ x^{2}$ climbs very fast once $∣ x ∣$ passes $2$ , so the branches hug near-vertical lines.

Look at the figure: the tall amber box drives the dashed asymptotes toward vertical, and the cyan branches fan open far wider than in Example 5.

Verify: At $b = 20$ : slope $= 10$ , $e = 1 + 100 = 101 \approx 10.05$ — large, as forecast. ✓ Together with Example 5 this brackets the whole range $1 < e < \infty$ . ✓

Worked example Example 7 — Case G: rectangular hyperbola

$a = b$ Show $\frac{x ^{2}}{25} - \frac{y ^{2}}{25} = 1$ is rectangular, and find its $e$ and asymptote angle.

Forecast: Equal denominators ⇒ $a = b$ ⇒ guess perpendicular ( $9 0^{\circ}$ apart) asymptotes and the signature $e = 2$ .

$a^{2} = b^{2} = 25 \Rightarrow a = b = 5$ . Why this step? Equal denominators force $a = b$ , the definition of a rectangular hyperbola.
Asymptotes $y = \pm \frac{b}{a} x = \pm x$ , i.e. at $\pm 4 5^{\circ}$ . Why? Slope $b / a = 1$ , and $tan 4 5^{\circ} = 1$ ; the two lines meet at $9 0^{\circ}$ .
$e = 1 + b^{2} / a^{2} = 1 + 1 = 2 \approx 1.414$ . Why? Ratio $b / a = 1$ plugged into $e = 1 + (b / a)^{2}$ .

Look at the figure: the box is a perfect square, so its diagonals are exactly perpendicular.

Verify: $e = 2 \approx 1.41421 > 1$ . ✓ Asymptote slopes $+ 1$ and $- 1$ multiply to $- 1$ , the perpendicularity test. ✓ Rotating $4 5^{\circ}$ turns this into an $x y = k$ form — see Rectangular Hyperbola xy=c^2.

Worked example Example 8 — Case H: word problem (LORAN navigation)

Two radio beacons $F_{1}$ and $F_{2}$ stand $200 km$ apart on an east–west line. A ship measures that its signal timing gives $P F_{1} - P F_{2} = 120 km$ (constant). On which curve does the ship lie? Give its equation with the beacons' midpoint as origin.

Forecast: "Constant difference of distances to two fixed points" is the very definition of a hyperbola, so the ship's position curve should be one branch — and since $P F_{1} > P F_{2}$ the ship is closer to $F_{2}$ .

Place origin at the midpoint, beacons at $F_{1} (- 100, 0)$ , $F_{2} (100, 0)$ (km). Why this step? Symmetric placement puts the centre at the origin and gives the standard form.
$2 c = 200 \Rightarrow c = 100$ . Why? $2 c$ is the focal separation, exactly the beacon distance.
$2 a = 120 \Rightarrow a = 60$ . Why? $2 a$ is the fixed distance-difference in the locus definition $∣ P F_{1} - P F_{2} ∣ = 2 a$ .
$b^{2} = c^{2} - a^{2} = 10 0^{2} - 6 0^{2} = 10000 - 3600 = 6400$ . Why? Rearranged plus-rule $c^{2} = a^{2} + b^{2}$ .
Equation $\frac{x ^{2}}{3600} - \frac{y ^{2}}{6400} = 1$ ; the ship is on the branch with $x > 0$ . Why that branch? $P F_{1} - P F_{2} = + 120 > 0$ means the ship is farther from $F_{1}$ , hence closer to $F_{2} (100, 0)$ , so it lies on the right branch $x > 0$ .

Look at the figure: the two amber beacons sit on the $x$ -axis, and the highlighted cyan right branch is where every " $F_{1}$ is $120$ km farther" position lives.

Verify: $a < c$ (60<100) as required for a real hyperbola. ✓ $b^{2} = 6400 > 0$ . ✓ $e = c / a = 100/60 = 5/3 \approx 1.667 > 1$ . ✓ Units: all lengths in km, ratios dimensionless. ✓ This is exactly the LORAN principle: constant time-difference × light-speed = constant distance-difference.

Worked example Example 9 — Case I: exam twist, numbers hidden

A hyperbola with horizontal transverse axis has eccentricity $e = \frac{13}{5}$ and one focus at $(13, 0)$ . Find $a$ , $b$ , and the equation.

Forecast: The focus gives $c = 13$ ; with $e = 13/5$ the number $5$ smells like $a$ . Guess $a = 5$ , then $b$ from the plus-rule.

$c = 13$ (focus is at distance $c$ on the $x$ -axis). Why this step? For a horizontal hyperbola centred at the origin, foci are $(\pm c, 0)$ .
$e = c / a \Rightarrow a = c / e = 13 \div \frac{13}{5} = 5$ . Why? Invert $e = c / a$ ; this recovers the hidden $a$ .
$b^{2} = c^{2} - a^{2} = 169 - 25 = 144 \Rightarrow b = 12$ . Why? Plus-rule again, solving for $b$ .
Equation $\frac{x ^{2}}{25} - \frac{y ^{2}}{144} = 1$ ; asymptotes $y = \pm \frac{12}{5} x$ . Why? Substitute $a^{2} = 25, b^{2} = 144$ ; slope $= b / a = 12/5$ .

Look at the figure: the amber right triangle with legs $a = 5$ , $b = 12$ and hypotenuse $c = 13$ shows the whole $(5, 12, 13)$ Pythagorean triple that the exam hid inside the numbers.

Verify: $c^{2} = a^{2} + b^{2}$ : $169 = 25 + 144$ . ✓ $e = c / a = 13/5 = 2.6 > 1$ . ✓ $(5, 12, 13)$ is a Pythagorean triple, a classic exam-friendly hint. ✓

Recall One-line summary of the matrix

Every hyperbola task reduces to: identify the transverse axis, then run P–S–P–E–A; degenerate = RHS $0$ , limiting = push $b \to 0$ (pinch) or $b \to \infty$ (fan), rectangular = $a = b$ . ::: All ten cells are the same five-step ritual applied to different givens.

Related building blocks: Eccentricity — unified conic definition, Ellipse — standard forms, foci, eccentricity (contrast the minus-rule), Parabola — standard forms, focus, directrix (the $e = 1$ boundary), and Conic Sections from a Double Cone for where all this geometry is born.

Where e=1+(b/a)2​ comes from

The scenario matrix

Worked Examples

Where $e = 1 + (b / a)^{2}$ comes from