Hyperbola — standard forms, asymptotes, foci, eccentricity: Visual walkthrough

We are going to earn, in order, these words: distance, focus, the constant $2 a$ , the constant $c$ , the leftover $b$ , and the asymptote lines. Nothing before it is drawn.

Step 1 — The rule: keep the difference of two distances fixed

WHAT. We invent a walking rule. Instead of keeping the sum $P F_{1} + P F_{2}$ constant (that traces an ellipse), we keep the difference constant: $dist to left pin P F_{1} - dist to right pin P F_{2} = a fixed number we choose 2 a .$

WHY the bars $∣ ∣$ ? The bars mean "throw away the minus sign". On the left arc $F_{1}$ is the far pin, on the right arc $F_{2}$ is — so the raw subtraction flips sign between the two arcs. Taking the size $∣ \dots ∣$ lets one rule cover both arcs at once. We wrote it as $2 a$ (not $a$ ) purely so that later a factor of $2$ cancels neatly; $a$ will turn out to be a half-length.

PICTURE. Below: two pins, a wandering point $P$ , and the two ruler-strings to it. The rule says the two string lengths must always be exactly $2 a$ apart.

Step 2 — Choose smart coordinates

WHAT. Put $F_{1} = (- c, 0), F_{2} = (+ c, 0),$ where $c > 0$ is half the pin separation (the pins are $2 c$ apart). Let the walking point be $P = (x, y)$ .

WHY $c$ and not just "the distance"? Giving the half-separation its own letter $c$ lets every formula stay symmetric: the left pin is at $- c$ , the right at $+ c$ , and the centre (midpoint) is the origin. Naming it now means later the beautiful relation between $a$ , $b$ , $c$ can even be written.

PICTURE. The pins are pinned to specific coordinates; $P = (x, y)$ floats. The two dashed segments are the distances we are about to write with the ruler formula.

So our rule becomes, letter by letter: $P F_{1} (x + c)^{2} + y^{2} - P F_{2} (x - c)^{2} + y^{2} = \pm 2 a .$ The $\pm$ is the same $∣ \dots ∣$ from Step 1, now written out: $+ 2 a$ traces one arc, $- 2 a$ the other.

Step 3 — Peel off the square roots (two careful squarings)

WHAT (move one root aside). $(x + c)^{2} + y^{2} = \pm 2 a + (x - c)^{2} + y^{2} .$ Now only one root sits alone on the left. WHY: squaring a lone root cleanly deletes it; squaring a sum of two roots would breed a cross-term with another root — no progress.

WHAT (square once). Squaring both sides and expanding: $(x + c)^{2} + y^{2} = 4 a^{2} \pm 4 a (x - c)^{2} + y^{2} + (x - c)^{2} + y^{2} .$ The terms $x^{2} + c^{2} + y^{2}$ appear on both sides and vanish; only the cross-terms $\pm 2 c x$ survive: $4 c x = 4 a^{2} \pm 4 a (x - c)^{2} + y^{2} ⟹ no root c x - a^{2} = \pm a one root left (x - c)^{2} + y^{2} .$

WHAT (square again). One root remains, so square once more: $c^{2} x^{2} - 2 a^{2} c x + a^{4} = a^{2} [(x - c)^{2} + y^{2}] = a^{2} x^{2} - 2 a^{2} c x + a^{2} c^{2} + a^{2} y^{2} .$ The middle terms $- 2 a^{2} c x$ match and cancel. Gather the survivors: $(c^{2} - a^{2}) x^{2} - a^{2} y^{2} = a^{2} (c^{2} - a^{2}) .$

PICTURE. Think of each squaring as flattening a bumpy root-curve into a smooth polynomial — the figure shows the two "locked boxes" being opened one at a time until only $x$ , $y$ , $a$ , $c$ remain.

Step 4 — Name the leftover $b^{2} = c^{2} - a^{2}$

WHAT. Define $b^{2} = c^{2} - a^{2} (b > 0) .$ Substitute into $(c^{2} - a^{2}) x^{2} - a^{2} y^{2} = a^{2} (c^{2} - a^{2})$ : $b^{2} x^{2} - a^{2} y^{2} = a^{2} b^{2} .$ Divide every term by $a^{2} b^{2}$ : $how far along the axis \frac{x ^{2}}{a ^{2}} - how far off it \frac{y ^{2}}{b ^{2}} = 1.$

WHY name $b$ ? Two payoffs. (1) The equation turns spotless. (2) $b$ is about to reveal itself as a real length on the picture — the vertical half-size of a hidden rectangle that sets the asymptote slopes. Contrast this with the ellipse, where $b^{2} = a^{2} - c^{2}$ (a minus) because there the pins sit inside and $c < a$ .

PICTURE. The final curve with $a$ , $b$ , $c$ all marked as honest lengths on one diagram.

Step 5 — Where the curve is allowed to live (every case)

WHAT — the cases.

$∣ x ∣ < a$ (the gap in the middle): impossible. No point of the curve lives strictly between the vertices. This is the empty channel that splits the curve into two branches.
$x = a$ or $x = - a$ : forces $y = 0$ . These are the two vertices — the pinch points, one per branch.
$x > a$ : the right branch, opening toward $+ \infty$ .
$x < - a$ : the left branch, opening toward $- \infty$ .
$y$ can be anything: for each allowed $x$ there are two $y$ values $\pm b x^{2} / a^{2} - 1$ — the curve is mirror-symmetric top/bottom.

WHY this matters. It guarantees you never get blindsided: there is no point of the hyperbola in the forbidden strip, and every branch is unbounded upward and downward.

PICTURE. The forbidden strip $∣ x ∣ < a$ shaded out; the two allowed regions carry the two branches.

Step 6 — The straight-line guides (asymptotes) appear

WHAT. Solve for $y$ exactly: $y = \pm \frac{b}{a} x \to 1 as ∣ x ∣ \to \infty 1 - \frac{a ^{2}}{x ^{2}} .$ As $∣ x ∣ \to \infty$ the square root sinks to $1$ , leaving $y \to \pm \frac{b}{a} x .$

Each term's job:

$\frac{b}{a}$ = slope = rise $b$ over run $a$ = (half-conjugate)/(half-transverse).
the $\pm$ = the two crossing lines (one climbs, one descends).
the shrinking $1 - a^{2} / x^{2}$ = the tiny "sag" that makes the curve just below the line and closes to zero.

WHY they never touch. The vertical gap is $\frac{b}{a} x (1 - 1 - \frac{a ^{2}}{x ^{2}}) \approx \frac{b}{a} x \cdot \frac{a ^{2}}{2 x ^{2}} = \frac{ab}{2 x} x \to \infty 0,$ positive but vanishing: the curve hugs the line forever without meeting it.

PICTURE. The dashed central rectangle, its two diagonal lines extended, and the curve pressed against them out at the edges.

Step 7 — Degenerate & limiting cases (so nothing surprises you)

WHAT — the special settings:

$a = b$ (rectangular / equilateral): the box is a square, asymptotes $y = \pm x$ meet at $9 0^{\circ}$ , and $e = 1 + 1 = 2$ . Rotate $4 5^{\circ}$ and it becomes the friendly $xy=k$ curve.
$b \to 0$ (with $a$ fixed): then $c \to a$ and $e = \frac{c}{a} \to 1^{+}$ . The branches squeeze flat onto the $x$ -axis outside the vertices — the narrowest possible hyperbola, edging toward the parabola limit $e = 1$ (see Eccentricity — unified conic definition).
$b \to \infty$ (with $a$ fixed): slope $b / a \to \infty$ , $e = 1 + b^{2} / a^{2} \to \infty$ . The branches fan open toward two vertical walls — the widest case.
$c \to a^{+}$ i.e. $2 a \to 2 c$ : the walking rule's constant approaches the whole pin gap. The triangle degenerates and the "curve" collapses onto the axis segment beyond the pins — the boundary of existence.

WHY. These pin down the full menu: $e > 1$ always, but how far above $1$ tunes everything from a razor-thin hyperbola to a nearly-square-cornered one.

PICTURE. Three hyperbolas sharing the same $a$ with growing $b$ (rising $e$ ), overlaid to show the branches fanning out.

The one-picture summary

Everything at once: two pins $F_{1}, F_{2}$ at $(\pm c, 0)$ ; the walking rule $∣ P F_{1} - P F_{2} ∣ = 2 a$ ; vertices at $(\pm a, 0)$ ; the leftover length $b$ with $b^{2} = c^{2} - a^{2}$ ; the central box $(\pm a, \pm b)$ whose diagonals $y = \pm \frac{b}{a} x$ are the asymptotes; the two branches hugging them.

Recall Feynman retelling — the whole walkthrough in plain words

Push two pins into a board. Walk so that one pin is always a fixed number of steps closer than the other — never mind the total, just the difference, and call that difference $2 a$ . Because a triangle's two sides can't differ by more than the third, that difference is smaller than the pin gap $2 c$ , so the pins live outside where you can walk closest. Trace all your legal standing spots: two swooping arcs, one hugging each pin, with a forbidden strip of width $2 a$ between them. Write "distance" with Pythagoras, square twice to erase the roots, and out pops $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ , where the new length $b$ satisfies $b^{2} = c^{2} - a^{2}$ — the vertical half-side of a hidden rectangle. Draw that rectangle's diagonals and extend them: those are the two straight rulers the arcs race alongside forever without ever scraping. Make the rectangle square ( $a = b$ ) and the rulers cross at a right angle; that's the rectangular hyperbola. That's the entire story — one rule about a difference of distances, and a picture that never lies.

Recall

Why are there two branches, not one? ::: The rule uses $∣ P F_{1} - P F_{2} ∣$ ; the raw difference is $+ 2 a$ on one side, $- 2 a$ on the other, so each sign traces its own arc. Why is $c > a$ forced (not chosen)? ::: A triangle's side difference is less than the third side, so $2 a < 2 c$ . What are the asymptote lines and their slope? ::: $y = \pm \frac{b}{a} x$ , the diagonals of the box with corners $(\pm a, \pm b)$ . Where does the curve refuse to go? ::: The strip $∣ x ∣ < a$ ; there $\frac{x ^{2}}{a ^{2}} < 1$ but the equation needs $\geq 1$ . What is $b$ geometrically? ::: The vertical half-side of the central rectangle, with $b^{2} = c^{2} - a^{2}$ ; the curve never reaches $(0, \pm b)$ .

Related: Parabola — standard forms, focus, directrix · Conic Sections from a Double Cone · Hinglish version