Definition via focus, directrix, eccentricity e: Visual walkthrough

We assume nothing except: you can measure the straight-line gap between two dots, and you can drop a perpendicular from a dot to a line. Everything else is built here.

Step 1 — Two special objects: a dot and a line

WHAT. Draw a single point $F$ (call it the focus) and a straight line $L$ (call it the directrix). Keep the point off the line.

WHY. Every conic is a competition between "how far am I from this dot" versus "how far am I from this wall." To even start the competition we need one dot and one wall on the page. Nothing else exists yet — no axes, no formula.

PICTURE. Look at the red dot $F$ and the black line $L$ . The tiny gap between them (drawn as a bracket) is what makes the whole thing possible; if $F$ sat on $L$ the two distances would collapse and there'd be no shape.

Figure — Definition via focus, directrix, eccentricity e

Step 2 — Measuring both distances for a roaming point $P$

WHAT. Pick any point $P$ on the page. Measure two things:

$P F$ = the straight gap from $P$ to the focus,
$P M$ = the perpendicular drop from $P$ to the line (the shortest possible walk to the wall — you hit it at a right angle). The point where that drop lands on the line is $M$ , the foot of the perpendicular from $P$ to $L$ .

WHY. These two lengths are the only ingredients in the rule. Note the asymmetry: distance to a point is the ordinary distance-formula length; distance to a line means the shortest route, which is always the perpendicular. Confusing these two is the single most common wreck — see Distance formula and distance from a point to a line.

PICTURE. The red segment is $P F$ (dot to dot). The black segment $P M$ meets $L$ at a little right-angle square at $M$ — that square is your reminder that only the perpendicular counts. Slide $P$ up or down and $P M$ is always horizontal here, because $L$ is vertical.

Step 3 — The rule that names the curve

WHAT. Demand that the ratio of the two distances is a fixed number $e$ , the same for every point on the curve: $to the dot P F = e \cdot to the wall P M$

WHY. A curve is just "the set of points obeying one rule." This is the rule. The number $e$ is a dial:

$e < 1$ forces $P F < P M$ : you must always hug the focus tighter than the wall → the curve closes.
$e = 1$ forces $P F = P M$ : a perfect tie → the curve just fails to close.
$e > 1$ lets $P F > P M$ : you may drift far from the focus → the curve splits open.

PICTURE. Three sample points, each with its red-string ( $P F$ ) and black-rod ( $P M$ ) drawn to scale. Read off the ratio: it's the same $e$ every time, even though both lengths change. That constancy is the curve's identity.

Step 4 — Turn the picture into coordinates

WHAT. Drop axes onto the picture. Put the focus at the origin $F = (0, 0)$ and the directrix as the vertical line $x = - d$ (with $d > 0$ , so the wall sits a distance $d$ to the left of the focus). Let $P = (x, y)$ .

Now write each distance in symbols: $P F = gap from (x, y) to (0, 0) x^{2} + y^{2}, P M = horizontal gap between x and - d ∣ x + d ∣$

WHY. Pictures are wonderful for feeling the rule, but to solve for the curve we need arithmetic. Axes convert "distance" into formulas: the focus distance is the plain distance formula, and the wall distance is just how far $x$ is from $- d$ — the $y$ -coordinate is irrelevant because the wall is vertical (the foot $M$ has the same $y$ as $P$ , so only the $x$ -gap survives).

PICTURE. The red $P F$ is the hypotenuse of a right triangle with legs $x$ and $y$ (that's why it's $x^{2} + y^{2}$ ). The black $P M$ is the flat horizontal run from the wall at $x = - d$ across to $x$ . The absolute value $∣ x + d ∣$ keeps it positive whether $P$ is left or right of the wall.

Step 5 — Apply the rule and square away the mess

WHAT. Substitute both distances into $P F = e \cdot P M$ : $x^{2} + y^{2} = e ∣ x + d ∣$ Both sides are lengths (never negative), so squaring loses no information and destroys both the root and the absolute value: $was P F^{2} x^{2} + y^{2} = was (e P M)^{2} e^{2} (x + d)^{2}$

WHY. The square root and the $∣ \cdot ∣$ are only bookkeeping to keep distances positive. Because we already know both sides are $\geq 0$ , squaring is safe here — it cannot invent fake solutions (that only happens when one side could be negative). This is the one algebra move that unlocks the whole family.

PICTURE. Left panel: the raw equation with a root and bars — jagged, hard to read. Right panel: after squaring, two clean quadratics. The red highlight follows the $e^{2}$ factor, the term that is about to decide everything.

Step 6 — Expand and watch the $x^{2}$ coefficient

WHAT. Multiply out $e^{2} (x + d)^{2} = e^{2} x^{2} + 2 e^{2} d x + e^{2} d^{2}$ , then gather: $shape switch (1 - e^{2}) x^{2} - 2 e^{2} d x + y^{2} = e^{2} d^{2}$

WHY. Stare at the coefficient in front of $x^{2}$ : it is $1 - e^{2}$ . This single quantity is the master switch. Its sign decides whether the curve closes, opens, or balances on the edge — and its value depends only on $e$ , not on where $P$ is. That's how one rule produces four different shapes.

PICTURE. A number line for $1 - e^{2}$ . Colour-coded zones: positive when $e < 1$ (both squares same sign → closed), zero when $e = 1$ (the $x^{2}$ term vanishes), negative when $e > 1$ (squares fight → open). The red marker sits exactly at $e = 1$ , the tipping point.

Step 7 — Case A: $e = 1$ , the parabola (perfect tie), and where $4 a$ comes from

WHAT. Put $e = 1$ . The $(1 - e^{2}) x^{2}$ term dies ( $1 - 1 = 0$ ), leaving only one power of $x$ : $y^{2} = 2 d x + d^{2}$ This is a parabola, but its vertex is not at the origin — our focus was at the origin, not the vertex. Let us find the vertex and re-centre.

Define the vertex. On the $x$ -axis ( $y = 0$ ) the equation gives $0 = 2 d x + d^{2}$ , so $x = - \frac{d}{2}$ . That point $V = (- \frac{d}{2}, 0)$ is the vertex — it sits halfway between the focus $(0, 0)$ and the wall $x = - d$ , exactly as symmetry demands.

Change variables to put $V$ at the origin. Let $X = x + \frac{d}{2} ⟺ x = X - \frac{d}{2} .$ This just slides the whole picture right by $\frac{d}{2}$ so the vertex becomes the new origin. Substitute into $y^{2} = 2 d x + d^{2}$ : $y^{2} = 2 d (X - \frac{d}{2}) + d^{2} = 2 d X - d^{2} + d^{2} = 2 d X .$

Define $a$ . In the new frame the focus lands at $X = \frac{d}{2}$ , so the vertex-to-focus distance is $a = \frac{d}{2}$ , i.e. $d = 2 a$ . Substituting $2 d = 4 a$ : $y^{2} = 4 a X$ (and dropping the temporary capital, $y^{2} = 4 a x$ in vertex-centred coordinates).

WHY. The mysterious $4$ is now fully earned: $a$ is the vertex-to-focus distance, the wall sits the same distance $a$ on the other side, and $2 d = 2 (2 a) = 4 a$ bundles those two equal gaps together. With no $x^{2}$ term, $y$ grows like $x$ forever — the U-shape that just barely fails to close. Full treatment: Parabola - Standard equation y^2=4ax.

PICTURE. The red parabola, vertex $V$ at the new origin, focus at $(a, 0)$ , directrix $x = - a$ . One sample point shows its red $P F$ equal in length to its black $P M$ — the defining tie $e = 1$ made visible.

Step 8 — Cases B & C: ellipse ( $e < 1$ ) and hyperbola ( $e > 1$ )

WHAT. Keep the general equation and read the two survivors of the switch:

$e < 1$ : $(1 - e^{2}) > 0$ . Both $x^{2}$ and $y^{2}$ carry positive coefficients → a bounded, closed ellipse.
$e > 1$ : $(1 - e^{2}) < 0$ . The $x^{2}$ and $y^{2}$ coefficients have opposite signs → an unbounded hyperbola with two branches.

WHY. When both squared terms push the same way, the point is trapped between them — you can't run to infinity without one term blowing up the balance, so the curve loops shut (ellipse). When the signs oppose, one term can grow as long as the other grows to match — so $x$ and $y$ escape to infinity together, tearing the curve into two open branches (hyperbola). See Ellipse - Standard form and c^2 = a^2 - b^2 and Hyperbola - Standard form and c^2 = a^2 + b^2.

PICTURE. Left: a red ellipse, both terms positive, closed loop. Right: a red hyperbola, opposite signs, two branches with dashed asymptotes. The tiny sign badges $(+, +)$ vs $(+, -)$ label exactly which coefficients cause which shape.

Step 9 — The degenerate edge: $e \to 0$ , the circle (with the algebra carried through)

WHAT. Setting $e = 0$ literally in the rule gives $P F = 0 \cdot P M = 0$ , collapsing the curve to the single point $F$ — useless. The honest limit keeps a real curve alive by letting the wall recede while $e$ shrinks, holding the product $e d$ fixed.

Introduce $R$ . Define the constant $R = e d (a fixed positive length) .$ Then $e = R / d$ . Feed this into the squared equation $x^{2} + y^{2} = e^{2} (x + d)^{2}$ : $x^{2} + y^{2} = \frac{R ^{2}}{d ^{2}} (x + d)^{2} = R^{2} (\frac{x}{d} + 1)^{2} .$

Take the limit $d \to \infty$ with $R$ fixed (this is exactly "the directrix recedes to infinity" while $e = R / d \to 0$ ). As $d \to \infty$ the term $\frac{x}{d} \to 0$ , so $(\frac{x}{d} + 1)^{2} \to 1$ and: $x^{2} + y^{2} = R^{2}$

WHY. That is the equation of a circle of radius $R$ centred at the focus $F = (0, 0)$ — so in the limit the focus becomes the centre and every point sits the same distance $R$ from it, with no directrix left in sight. The circle is genuinely the $e \to 0$ member of the same family, and now we have carried its formula through on-page, not just its picture. See Circle as limiting case e=0.

PICTURE. A sequence: an ellipse at $e = 0.6$ , a rounder one at $e = 0.3$ , and a perfect red circle $x^{2} + y^{2} = R^{2}$ at $e \to 0$ , with the directrix arrow sliding off the right edge of the frame to signal "wall gone to infinity."

The one-picture summary

Everything on one dial. Below: the focus $F$ (red) fixed, the directrix $L$ fixed, and the same rule $P F / P M = e$ drawing four curves as $e$ climbs. Watch the $(1 - e^{2})$ switch flip from positive (ellipse) through zero (parabola) to negative (hyperbola), with the circle $x^{2} + y^{2} = R^{2}$ waiting at $e \to 0$ .

Recall Feynman retelling — the whole walkthrough in plain words

Put a dot on the floor and paint a straight line nearby. Now walk around dropping pebbles, but only where your steps to the dot are a fixed multiple $e$ of your steps to the line. The nearest point on the line to you — where a straight-across walk lands — is the foot $M$ .

If $e$ is small (say a half) you must always stay closer to the dot, so your pebbles curl into a closed oval — an ellipse. If $e$ is exactly one, your two walks tie, and the pebbles make a U that opens forever but never closes — a parabola. If $e$ is bigger than one you may wander far from the dot, and the pebbles split into two curves flying apart — a hyperbola. If $e$ shrinks all the way to zero while the line marches off to infinity, its pull flattens out and every pebble sits the same distance $R = e d$ from the dot — a circle $x^{2} + y^{2} = R^{2}$ .

To find the equation we just wrote both distances with coordinates ( $x^{2} + y^{2}$ to the dot, $∣ x + d ∣$ to the wall), set their ratio to $e$ , and squared. Out popped one master equation whose $x^{2}$ -coefficient, $1 - e^{2}$ , secretly decides the shape: positive closes it, zero straightens it, negative splits it. For the parabola we slid the vertex to the origin and found $2 d = 4 a$ ; for the circle we froze $R = e d$ and let the wall run away. One dial, four curves.

Recall check

Why does squaring not create fake solutions here?

Both sides (

P F

and

e P M

) are non-negative distances, so squaring is reversible.

What is the value of

(1 - e^{2})

for a parabola?

Zero — it kills the

x^{2}

term, leaving

y^{2} = 4 a x

after re-centring.

Where is the parabola's vertex before re-centring?

x = - d /2

, halfway between focus

(0, 0)

and directrix

x = - d

How does the

4 a

appear?

With vertex-to-focus distance

a = d /2

, we get

d = 2 a

, so

2 d = 4 a

Same-sign squared coefficients give which conic?

An ellipse (bounded, closed).

Opposite-sign squared coefficients give which conic?

A hyperbola (two open branches).

What is held fixed as

e \to 0

and what results?

R = e d

is held fixed; the limit gives the circle

x^{2} + y^{2} = R^{2}

Connections

Parent — Focus, directrix, eccentricity
Parabola - Standard equation y^2=4ax
Ellipse - Standard form and c^2 = a^2 - b^2
Hyperbola - Standard form and c^2 = a^2 + b^2
Circle as limiting case e=0
Distance formula and distance from a point to a line
Latus rectum and semi-latus rectum