Hyperbola — standard forms, asymptotes, foci, eccentricity: Exercises

Before we start, a one-line reminder of the four numbers that describe a hyperbola, all pinned to the same picture so no symbol is a mystery:

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

Look at the picture: the two branches (magenta) open left–right; the dashed violet rectangle has corners at $(\pm a, \pm b)$ ; its diagonals extended (orange) are the asymptotes the curve hugs forever; the orange dots are the foci at $(\pm c, 0)$ , sitting outside the vertices.

Level 1 — Recognition

Goal: read a hyperbola's parts straight off the equation, no heavy algebra.

Recall Solution — L1·Q1

WHAT to spot: the term with the plus sign tells you the transverse direction. Here $\frac{x ^{2}}{9}$ carries the $+$ , so the transverse axis is along the == $x$ -axis==. Read the denominators: the number under the $+$ term is $a^{2}$ , the other is $b^{2}$ .

$a^{2} = 9 \Rightarrow a = 3$ .
$b^{2} = 16 \Rightarrow b = 4$ . Vertices sit at $(\pm a, 0) = (\pm 3, 0)$ — the tips of the two branches, on the transverse axis.

Recall Solution — L1·Q2

The tell: a minus sign between the two squared terms $\Rightarrow$ hyperbola. A plus sign $\Rightarrow$ ellipse.

(A) has $+$ → ellipse, not a hyperbola.
(B) has $-$ → hyperbola. The $+$ term is $\frac{y ^{2}}{25}$ , so the transverse axis is along the $y$ -axis, meaning the branches open up and down.

Level 2 — Application

Goal: plug into $c^{2} = a^{2} + b^{2}$ , $e = c / a$ , and the asymptote slope.

Recall Solution — L2·Q1

$a^{2} = 16 \Rightarrow a = 4$ , and $b^{2} = 9 \Rightarrow b = 3$ .
Focus relation (WHY plus): the foci sit outside the vertices, so they are farther from the centre than $a$ ; the relation is $c^{2} = a^{2} + b^{2}$ . Thus $c^{2} = 16 + 9 = 25 \Rightarrow c = 5$ . Foci $(\pm 5, 0)$ .
$e = \frac{c}{a} = \frac{5}{4} = 1.25$ . Sanity check: $e > 1$ ✓ (a hyperbola always has $e > 1$ ).
Asymptote slope $= \frac{b}{a} = \frac{3}{4}$ (rise $b$ over run $a$ — the corner of the box is at $(a, b) = (4, 3)$ ). Asymptotes: $y = \pm \frac{3}{4} x$ .

Recall Solution — L2·Q2

First, orientation. The $+$ term is $\frac{y ^{2}}{36}$ , so this is the vertical form $\frac{y ^{2}}{a ^{2}} - \frac{x ^{2}}{b ^{2}} = 1$ : transverse axis on the $y$ -axis, foci on the $y$ -axis.

$a^{2} = 36 \Rightarrow a = 6$ (under the $+$ term). $b^{2} = 64 \Rightarrow b = 8$ .
$c^{2} = a^{2} + b^{2} = 36 + 64 = 100 \Rightarrow c = 10$ . Foci $(0, \pm 10)$ .
Asymptote slope for the vertical form. The box now stretches to $y = \pm a$ , $x = \pm b$ , so its corner is at $(b, a) = (8, 6)$ and the slope is $\frac{a}{b} = \frac{6}{8} = \frac{3}{4}$ . Asymptotes: $y = \pm \frac{6}{8} x = \pm \frac{3}{4} x$ .

Level 3 — Analysis

Goal: reason backwards from $e$ , from angles, from limiting behaviour.

Recall Solution — L3·Q1

Why tangent enters: the asymptote is a straight line through the origin; the slope of a line at angle $θ$ to the $x$ -axis is $tan θ$ — "rise over run" is exactly what $tan$ measures. So slope $= tan 3 0^{\circ} = \frac{1}{3}$ . Connect slope to shape: for the horizontal form, slope $= \frac{b}{a} = \frac{1}{3}$ , so $\frac{b ^{2}}{a ^{2}} = \frac{1}{3}$ . Use the shortcut $e = 1 + \frac{b ^{2}}{a ^{2}}$ (this comes from $c^{2} = a^{2} + b^{2}$ divided by $a^{2}$ ), which lets us find $e$ without ever knowing $a$ or $b$ separately: $e = 1 + \frac{1}{3} = \frac{4}{3} = \frac{2}{3} \approx 1.155.$ $e > 1$ ✓.

Recall Solution — L3·Q2

Foci at $(\pm 5, 0)$ ⇒ $c = 5$ , and transverse axis on the $x$ -axis.
$e = \frac{c}{a} \Rightarrow \frac{5}{3} = \frac{5}{a} \Rightarrow a = 3$ . (Rearranged: $a = \frac{c}{e}$ .)
$b^{2} = c^{2} - a^{2} = 25 - 9 = 16$ (same $c^{2} = a^{2} + b^{2}$ , just solved for $b^{2}$ ). So $b = 4$ .
Equation: $\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1$ .

Recall Solution — L3·Q3

Limiting reasoning: $e = 1 + \frac{b ^{2}}{a ^{2}}$ . As $b \to \infty$ the fraction $\frac{b ^{2}}{a ^{2}} \to \infty$ , so $e \to \infty$ . What it looks like: the asymptote slope $\frac{b}{a} \to \infty$ means the asymptotes tilt toward vertical. The branches fan open until they run almost parallel to the $y$ -axis — the curve spreads as wide as possible. So big $b$ ⇒ big $e$ ⇒ maximally open branches.

Level 4 — Synthesis

Goal: assemble the equation from mixed, indirect clues.

Recall Solution — L4·Q1

Clue 1 (asymptote): slope $\frac{b}{a} = \frac{2}{3}$ , so $b = \frac{2}{3} a$ , i.e. $b^{2} = \frac{4}{9} a^{2}$ . Clue 2 (point on curve): substitute $(6, 4)$ into $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ : $\frac{36}{a ^{2}} - \frac{16}{b ^{2}} = 1.$ Replace $b^{2} = \frac{4}{9} a^{2}$ : $\frac{36}{a ^{2}} - \frac{16}{\frac{4}{9} a ^{2}} = \frac{36}{a ^{2}} - \frac{36}{a ^{2}} = 0 \neq = 1.$ Uh-oh — that gives $0$ , not $1$ . The point $(6, 4)$ lies exactly on the asymptote ( $4 = \frac{2}{3} \cdot 6$ ), so no hyperbola with that asymptote can pass through it. This is a trap-check: the data is inconsistent. (See the L4 mistake box.) The honest answer is: no such hyperbola exists.

Recall Solution — L4·Q2

Clue 1: $\frac{b}{a} = \frac{1}{2} \Rightarrow b^{2} = \frac{1}{4} a^{2}$ . Clue 2: plug $(6, 4)$ into $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ : $\frac{36}{a ^{2}} - \frac{16}{\frac{1}{4} a ^{2}} = \frac{36}{a ^{2}} - \frac{64}{a ^{2}} = \frac{- 28}{a ^{2}} .$ This equals $- 28/ a^{2}$ , which is negative, never $1$ . That signals the point lies in the region "below" the horizontal branches — it actually belongs to a conjugate (vertical) hyperbola with the same asymptotes, of form $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = - 1$ . Solving with $= - 1$ : $\frac{- 28}{a ^{2}} = - 1 \Rightarrow a^{2} = 28, b^{2} = \frac{1}{4} a^{2} = 7.$ Equation: $\frac{x ^{2}}{28} - \frac{y ^{2}}{7} = - 1$ , equivalently $\frac{y ^{2}}{7} - \frac{x ^{2}}{28} = 1$ . Check: $\frac{16}{7} - \frac{36}{28} = \frac{16}{7} - \frac{9}{7} = 1$ ✓.

Recall Solution — L4·Q3

Foci $(\pm 13, 0)$ ⇒ $c = 13$ , transverse on $x$ -axis.
Transverse length $= 2 a = 10 \Rightarrow a = 5$ .
$b^{2} = c^{2} - a^{2} = 169 - 25 = 144 \Rightarrow b = 12$ .
Equation: $\frac{x ^{2}}{25} - \frac{y ^{2}}{144} = 1$ . (Also $e = c / a = 13/5 = 2.6$ .)

Level 5 — Mastery

Goal: full multi-step problems, degenerate cases, and the rectangular hyperbola.

Recall Solution — L5·Q1

Recall the directrix: for the horizontal form the directrices are $x = \pm \frac{a}{e}$ . So $\frac{a}{e} = 2$ .

With $e = 2$ : $a = 2 e = 2 \cdot 2 = 4$ .
$c = e a = 2 \cdot 4 = 8$ .
$b^{2} = c^{2} - a^{2} = 64 - 16 = 48$ .
Equation: $\frac{x ^{2}}{16} - \frac{y ^{2}}{48} = 1$ .

Recall Solution — L5·Q2

What "rectangular" gives us: $a = b$ , so the guiding box is a square, its diagonals are perpendicular, and $e = 1 + \frac{b ^{2}}{a ^{2}} = 1 + 1 = 2$ . (Every rectangular hyperbola has $e = 2$ .)

$c = e a = 2 a$ . We're told $c = 42$ , so $a = 4$ , hence $b = 4$ .
Check $c$ directly: $c^{2} = a^{2} + b^{2} = 16 + 16 = 32$ , and $32 = 42$ ✓.
Asymptotes: slope $\frac{b}{a} = 1$ , so $y = \pm x$ (perpendicular guides).

Recall Solution — L5·Q3

This is the raw definition. The locus $∣ P F_{1} - P F_{2} ∣ = 2 a$ is a hyperbola with the foci at $(\pm c, 0)$ .

Foci $(\pm 5, 0)$ ⇒ $c = 5$ .
Constant difference $= 8 = 2 a \Rightarrow a = 4$ .
Degeneracy check: we need $2 a < 2 c$ , i.e. $a < c$ : here $4 < 5$ ✓ (if $a$ equalled $c$ the "curve" would collapse to two rays; if $a > c$ nothing exists).
$b^{2} = c^{2} - a^{2} = 25 - 16 = 9$ .
Equation: $\frac{x ^{2}}{16} - \frac{y ^{2}}{9} = 1$ , with $e = c / a = 5/4 = 1.25$ .

Recall Solution — L5·Q4

The degenerate case. Setting the right side to $0$ factors: $\frac{x ^{2}}{16} - \frac{y ^{2}}{9} = 0 \Rightarrow (\frac{x}{4} - \frac{y}{3}) (\frac{x}{4} + \frac{y}{3}) = 0 \Rightarrow y = \pm \frac{3}{4} x .$ So $= 0$ gives exactly the pair of asymptotes — the two crossing straight lines. This is the "hyperbola with its $1$ removed," which is why far from the centre (where the $1$ becomes negligible) the real curve of L5·Q3 hugs these very lines. The $= 0$ equation is the skeleton the $= 1$ curve wraps around.

Recall One-screen recap of every formula used here

Standard forms ::: $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ (horizontal) and $\frac{y ^{2}}{a ^{2}} - \frac{x ^{2}}{b ^{2}} = 1$ (vertical) Focus relation ::: $c^{2} = a^{2} + b^{2}$ (foci outside vertices) Eccentricity ::: $e = \frac{c}{a} = 1 + \frac{b ^{2}}{a ^{2}} > 1$ Asymptote slope (horizontal form) ::: $\pm \frac{b}{a}$ ; (vertical form) $\pm \frac{a}{b}$ Directrices (horizontal) ::: $x = \pm \frac{a}{e}$ Defining locus ::: $∣ P F_{1} - P F_{2} ∣ = 2 a$ , requiring $a < c$ Rectangular hyperbola ::: $a = b$ , $e = 2$ , asymptotes $y = \pm x$ $= 0$ instead of $= 1$ ::: gives the asymptote pair (degenerate hyperbola)

Related vault pages: Ellipse — standard forms, foci, eccentricity, Parabola — standard forms, focus, directrix, Eccentricity — unified conic definition, Asymptotes and Limits at Infinity, Rectangular Hyperbola xy=c^2, Conic Sections from a Double Cone.