Visual walkthrough — Ellipse — standard forms, semi-major - minor axes, foci, eccentricity, latus rectum

We assume nothing except that you can measure the straight-line distance between two dots. We even build that from scratch in Step 0.

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Step 0 — What "distance" means (the one tool we need)

WHAT. Two dots on paper. The straight-line gap between dot $A=(x_1,y_1)$ and dot $B=(x_2,y_2)$.

WHY this tool. The whole ellipse is defined by adding two distances. So before anything else we must know how to turn two dots into a number. The only tool that does this from coordinates is the Pythagorean theorem.

PICTURE.

Look at the amber right triangle. Its horizontal leg is how far apart the dots are left-to-right, $\Delta x=x_2-x_1$. Its vertical leg is the up-down gap, $\Delta y=y_2-y_1$. The hypotenuse — the slanted cyan line — is the actual distance.

$\underset{\text{the gap we want}}{\underbrace{d}}=\sqrt{\underset{\text{horizontal leg squared}}{\underbrace{(\Delta x{)}^2}}+\underset{\text{vertical leg squared}}{\underbrace{(\Delta y{)}^2}}}$

The square root is there because Pythagoras gives us $d^2$; we want $d$, so we undo the square.

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Step 1 — Pin the two foci and stretch the string

WHAT. We place two special points — the foci (nails) — and demand that a moving pencil point $P$ keeps the total string length fixed.

WHY put them on the $x$-axis, symmetric about the origin? We are free to choose where to draw our grid. Choosing the foci at $(-c,0)$ and $(+c,0)$ makes the picture symmetric left-right and up-down, which will make almost everything cancel later. Laziness now = clean algebra later.

PICTURE.

• The two amber dots are the foci $F_1=(-c,0)$ and $F_2=(+c,0)$. Here $c>0$ is just half the distance between the nails.
• The cyan point $P=(x,y)$ is the pencil, free to roam.
• The two white strings are the distances $P{F}_1$ and $P{F}_2$.

The ellipse rule (the string never changes length): $\underset{\text{dist to left nail}}{\underbrace{P{F}_1}}+\underset{\text{dist to right nail}}{\underbrace{P{F}_2}}=\underset{\text{fixed string length}}{\underbrace{2a}}$

We name the constant $2a$ (not just "$k$") because it will turn out to be the full width of the finished oval — naming it cleverly now saves relabelling later.

Now feed in Step 0's distance formula for each string: $\sqrt{\underset{\text{from }x-(-c)}{\underbrace{(x+c{)}^2}}+y^2}+\sqrt{\underset{\text{from }x-(c)}{\underbrace{(x-c{)}^2}}+y^2}=2a$

That is the ellipse — fully. Everything below is just tidying this into a shape we can read.

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Step 2 — Get one square root alone

WHAT. Move the second root to the right so ONE root sits by itself.

WHY. Squaring a sum of two roots leaves a root behind (the cross-term $2\sqrt{}\sqrt{}$). Squaring a lone root kills it outright. So we isolate first.

PICTURE.

Think of the equation as a balance scale. The left pan holds root $R_1$; we slide root $R_2$ over to the right pan — the scale stays level because we did the same thing to both sides.

$\underset{R_1,\text{now alone}}{\underbrace{\sqrt{(x+c{)}^2+y^2}}}=\underset{\text{constant}}{\underbrace{2a}}-\underset{R_2,\text{moved over}}{\underbrace{\sqrt{(x-c{)}^2+y^2}}}$

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Step 3 — Square once, watch the twins cancel

WHAT. Square both sides. Then expand and cancel every term that appears identically on both sides.

WHY. Squaring the left destroys $R_1$. On the right we get $4a^2-4a{R}_2+R_2^2$ — still one root, but now only one, which we can chase down next.

PICTURE.

The figure colours matching pieces the same: $y^2$ (cyan) appears on both sides, and so do $x^2$ and $c^2$ (white). Same colour, both sides ⇒ they annihilate.

Squaring: $(x+c{)}^2+y^2=4a^2-4a\sqrt{(x-c{)}^2+y^2}+(x-c{)}^2+y^2$

Expand the two brackets: $x^2+\underset{\text{keep}}{\underbrace{2cx}}+c^2+y^2=4a^2-4a\sqrt{}+x^2\underset{\text{keep}}{\underbrace{-2cx}}+c^2+y^2$

Cancel the twins $x^2,c^2,y^2$ from both sides. What survives: $2cx=4a^2-4a\sqrt{(x-c{)}^2+y^2}-2cx⟹4cx=4a^2-4a\sqrt{(x-c{)}^2+y^2}$

Divide by $4$ and tidy so the lone root is alone again: $a\sqrt{(x-c{)}^2+y^2}=\underset{\text{no roots here}}{\underbrace{a^2-cx}}$

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Step 4 — Square the second (last) time

WHAT. Square both sides once more. Now zero roots remain — it becomes ordinary algebra.

WHY. One root left ⇒ one more squaring finishes the job. This is why the recipe is "isolate, square, isolate, square": exactly two roots need exactly two squarings.

PICTURE.

$a^2[(x-c{)}^2+y^2]=(a^2-cx{)}^2$

Expand both sides fully: $\underbrace{a^2{x}^2}\underset{\text{will cancel}}{\underbrace{-2a^{2}cx}}+a^2{c}^2+a^2{y}^2=a^4\underset{\text{will cancel}}{\underbrace{-2a^{2}cx}}+c^2{x}^2$

The amber term $-2a^{2}cx$ sits on both sides — cancel it. Then gather every $x^2$ on the left and constants on the right: $a^2{x}^2-c^2{x}^2+a^2{y}^2=a^4-a^2{c}^2$

Factor each side: $\underset{\text{group}}{\underbrace{(a^2-c^2)}}{x}^2+a^2{y}^2=a^2\underset{\text{same group}}{\underbrace{(a^2-c^2)}}$

The same bundle $(a^2-c^2)$ appears twice. That is a signal we should give it a name.

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Step 5 — Name the bundle $b^2=a^2-c^2$ (and prove it is positive)

WHAT. Define a new length $b$ by $b^2=a^2-c^2$.

WHY it must be positive (so $b$ is a real length). The string $2a$ is looped over the nails and pulled taut past them — it is longer than the straight nail-to-nail gap $2c$. Longer string ⇒ $2a>2c$ ⇒ $a>c$ ⇒ $a^2-c^2>0$. A positive number always has a real square root, so $b$ genuinely exists.

PICTURE.

This is the famous hidden right triangle: from the centre, go up a distance $b$ to a co-vertex and out a distance $c$ to a focus; the slanted side back to a vertex has length $a$. So $a^2=b^2+c^2$ — it is the Pythagorean theorem living inside the ellipse.

Substituting $b^2$: $b^2{x}^2+a^2{y}^2=a^2{b}^2$

Divide every term by $a^2{b}^2$ to make the right side $1$: $\frac{b^2{x}^2}{a^2{b}^2}+\frac{a^2{y}^2}{a^2{b}^2}=\frac{a^2{b}^2}{a^2{b}^2}$

$\boxed{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1}(a>b>0)$

Each symbol earned: $a$ = half the string = distance centre-to-vertex; $b=\sqrt{a^2-c^2}$ = distance centre-to-co-vertex; $c$ = half the nail gap = distance centre-to-focus.

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Step 6 — Read the shape back off the equation

WHAT. Plug in the two easy cases to find where the curve crosses the axes.

WHY. An equation is only useful if we can see it. Setting one variable to zero finds the crossings — the skeleton of the oval.

PICTURE.

• Set $y=0$: $\frac{x^2}{a^2}=1\Rightarrow x=\pm a$. Vertices $(\pm a,0)$ — the widest points.
• Set $x=0$: $\frac{y^2}{b^2}=1\Rightarrow y=\pm b$. Co-vertices $(0,\pm b)$ — the narrowest points.

Because $a>b$, the $x$-direction is the long one: the major axis (length $2a$) is horizontal, the minor axis (length $2b$) vertical, and the foci $(\pm c,0)$ sit on the major axis, tucked inside the vertices since $c<a$.

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Step 7 — The degenerate & limiting cases (nothing left uncovered)

WHAT. Push the construction to its extremes and see what the same formula produces.

WHY. A derivation you trust must survive its boundaries. Three edge cases:

PICTURE.

1. Nails coincide ($c=0$). Then $b^2=a^2-0=a^2$, so $b=a$, and the equation becomes $\frac{x^2}{a^2}+\frac{y^2}{a^2}=1$, i.e. $x^2+y^2=a^2$ — a Circle of radius $a$. Eccentricity $e=c/a=0$. A circle is a "lazy ellipse."
2. Nails race to the vertices ($c\to a$). Then $b^2=a^2-c^2\to 0$, so $b\to 0$: the oval flattens toward the segment between the foci. Here $e=c/a\to 1$ — maximally squashed but never reaching $1$ (at $e=1$ the shape opens up into a Parabola, a different beast).
3. Bigger denominator under $y^2$. If we had pinned the nails on the $y$-axis instead, the identical algebra gives $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$ with foci $(0,\pm c)$ — a vertical major axis. Same derivation, axes swapped.

Contrast: the Hyperbola — standard forms uses the difference of the two strings, flipping a sign to $c^2=a^2+b^2$. Same skeleton, opposite temperament — see Eccentricity and directrix for the unifying view, and the Conic Sections — overview for where all four curves come from one cone.

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The one-picture summary

Everything at once: the two nails at $(\pm c,0)$, a taut string of total length $2a$ pinning a point $P$, the vertices at $(\pm a,0)$, co-vertices at $(0,\pm b)$, the right triangle $a^2=b^2+c^2$ standing from the centre, the latus-rectum chord of length $2b^2/a$ across a focus, and the finished equation floating above the oval.

Recall Feynman: the whole walkthrough in plain words

Hammer two nails a little apart. Drop a string loop over them and pull it tight with a pencil — the pencil always keeps the two nail-distances adding to the same total, the string length, which we call $2a$. Write "distance to left nail plus distance to right nail equals $2a$" using Pythagoras for each distance, and you get an equation with two ugly square roots. You can't kill both roots at once, so you shove one root to the far side, square (one root dies, the twin terms on both sides cancel), then shove the surviving root aside and square again (the last root dies). What's left factors and keeps showing the bundle $a^2-c^2$; because the string is longer than the nail gap, that bundle is positive, so we name it $b^2$. Divide through and — poof — $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Setting $y=0$ finds the far ends ($\pm a$), $x=0$ finds the near ends ($\pm b$), and the little right triangle with legs $b$ and $c$ and hypotenuse $a$ ties them together. Slide the nails together and you get a circle; slide them to the ends and the oval goes flat. That's the entire story.

Connections

• Parent topic
• Pythagorean theorem — the distance formula and the internal $a^2=b^2+c^2$ triangle
• Circle — the $c=0$, $e=0$ degenerate case
• Parabola — the $e=1$ limiting boundary
• Hyperbola — standard forms — same method, difference of distances, $c^2=a^2+b^2$
• Eccentricity and directrix — the unified focus–directrix viewpoint
• Conic Sections — overview — all conics as slices of one cone