Kepler's connection — orbits are ellipses (motivation): Visual walkthrough

We assume you know nothing yet — not what $a$ , $b$ , $c$ , $e$ , "focus", or "eccentricity" mean. Each becomes a picture before it becomes a letter.

Step 1 — Two pins and a loop of string (what a focus is)

WHAT. Push two pins into a board. Loop a piece of string around them, pull it taut with a pencil, and drag the pencil all the way around. The curve you get is an ellipse.

WHY. This is the honest, physical definition — no formula yet. Each pin is called a focus (plural foci). The string never changes length, so no matter where the pencil is, the two straight bits — from pencil to pin-1 and from pencil to pin-2 — always add up to the same total. That constant total is the one rule the whole shape obeys.

PICTURE. In the figure below, the pencil sits at a point $P$ . The two coloured segments are its distances to the pins, $r_{1}$ (magenta) and $r_{2}$ (violet). The claim is $r_{1} + r_{2}$ is the same for every point on the curve.

Figure — Kepler's connection — orbits are ellipses (motivation)

Step 2 — Put it on axes and read off the widest and tallest points

WHAT. Slide the whole picture so the midpoint between the two pins sits at the origin $(0, 0)$ , and lay the pins along the horizontal axis. Now the ellipse is symmetric left–right and up–down.

WHY. Symmetry is a gift: it turns "drag a pencil" into an equation we can compute with. With the pins on the $x$ -axis, the curve reaches farthest along $x$ (the widest points) and along $y$ (the tallest points). Those four special points are where all our lengths will get their names.

PICTURE. The rightmost point is called a vertex. The distance from the centre out to it we name $a$ — the semi-major axis (half of the long way across). The distance from the centre up to the top point we name $b$ — the semi-minor axis (half of the short way). The distance from the centre out to a pin we name $c$ .

Step 3 — Prove the string length is exactly $2 a$

WHAT. We show that the mystery constant "total string length" equals $2 a$ — twice the semi-major axis — so the name in Step 1 was justified.

WHY. We need a concrete value for the constant sum before any algebra works. The trick: evaluate the constant-sum rule at a cleverly chosen point where the two distances are easy to read.

PICTURE. Look at the right vertex, the point $(a, 0)$ . Its distance to the near pin (at $x = c$ ) is $a - c$ . Its distance to the far pin (at $x = - c$ ) is $a + c$ . Add them: $to near focus (a - c) + to far focus (a + c) = 2 a .$ The $c$ 's cancel — the sum is $2 a$ regardless of where the pins sit. So the constant string length is $2 a$ . That is why we wrote $2 a$ back in Step 1.

Step 4 — The right triangle that links $a$ , $b$ , $c$

WHAT. We find how $a$ , $b$ , $c$ relate. Answer: they form a right triangle, giving $c^{2} + b^{2} = a^{2}$ .

WHY. We used a point on the major axis in Step 3. Now use the top point $(0, b)$ instead — a different special point that pins down $b$ . This gives us a second equation, and two equations let us eliminate one unknown later.

PICTURE. At the top point $(0, b)$ , by up–down and left–right symmetry, the pencil is equally far from both pins. Since the two distances are equal and add to $2 a$ , each one must be exactly $a$ . Now drop a line from that top point straight down to the centre: length $b$ . From the centre out to a pin: length $c$ . The slanted string from top point to pin: length $a$ . That is a right-angled triangle with legs $b$ and $c$ and hypotenuse $a$ .

Step 5 — Name the "off-centre-ness": eccentricity $e$

WHAT. We define eccentricity $e = c / a$ : how far the pins sit from the centre, measured as a fraction of the big radius.

WHY. Absolute distances ( $c$ in km) don't tell you the shape — a huge circle and a tiny circle are the same shape. What tells you the shape is the ratio. Dividing $c$ by $a$ strips out the size and leaves pure shape. That is why we divide, and why we don't just quote $c$ .

PICTURE. Slide the pins from the centre outward and watch $e = c / a$ climb from $0$ to nearly $1$ :

$e = 0$ : pins together at the centre → a perfect circle.
small $e$ : pins barely off-centre → a nearly round ellipse (Earth, $e = 0.017$ ).
$e \to 1$ : pins near the vertices → a long thin cigar (comet, $e = 0.97$ ).

Now substitute $c = a e$ into the triangle $b^{2} = a^{2} - c^{2}$ : $b^{2} = a^{2} - (a e)^{2} = a^{2} (1 - e^{2}) .$ Each piece: $a^{2}$ is the big radius squared, $e^{2}$ scales down the correction, and the result is the small radius squared. This is the $b^{2} = a^{2} (1 - e^{2})$ your parent note stated — now earned.

Step 6 — Where the planet is nearest and farthest (perihelion / aphelion)

WHAT. Put the Sun at one focus (say the right pin, at $x = c = a e$ ). The planet's distance to the Sun is smallest at one vertex and largest at the other. We compute both.

WHY. Kepler's first law says the Sun lives at a focus, not the centre. So the interesting distances are focus-to-vertex, not centre-to-vertex. There are exactly two vertices, so exactly two extreme distances — the nearest and the farthest a planet ever gets.

PICTURE. The near vertex sits at $x = a$ ; the Sun sits at $x = a e$ . Their gap is $a - a e = a (1 - e)$ — the perihelion (closest). The far vertex is at $x = - a$ ; its gap to the Sun is $a + a e = a (1 + e)$ — the aphelion (farthest).

Step 7 — Degenerate & edge cases (never leave a gap)

WHAT. We check the boundary values of $e$ so no reader hits an unshown scenario.

WHY. A definition that only works for "nice" inputs is a trap. We must show $e = 0$ and $e \to 1$ explicitly.

PICTURE.

$e = 0$ (circle). Then $c = a e = 0$ : both pins land on the centre, the "string" makes a circle, and $r_{m i n} = r_{m a x} = a$ — nearest equals farthest, which is exactly what "constant radius" means.
$e \to 1$ (opening up). Then $b^{2} = a^{2} (1 - e^{2}) \to 0$ : the ellipse gets infinitely thin. At exactly $e = 1$ the curve stops closing — it becomes a parabola (see Parabola — projectile motion connection), and for $e > 1$ a hyperbola (an escaping orbit — see Hyperbola — unbound orbits and escape). The single number $e$ chooses the whole family, as Eccentricity of a conic and the Focus–directrix definition of conics make precise.

The one-picture summary

Every length, every name, every result on one diagram: the string of total length $2 a$ ; the focus triangle $b^{2} + c^{2} = a^{2}$ ; the Sun at focus $x = a e$ ; and the two extreme distances $a (1 \pm e)$ .

Recall Feynman retelling — the whole walkthrough in plain words

I pin down two thumbtacks and loop a string around them. Dragging a pencil keeps the string tight, so the two distances from my pencil to the tacks always add to the same amount — that's an ellipse, and each tack is a "focus." I slide the picture so the middle is at zero. Half the long way across I call $a$ ; half the short way I call $b$ ; the tack's distance from the middle I call $c$ . Standing at the far right corner, my two distances are $(a - c)$ and $(a + c)$ , which add to $2 a$ — so the string is $2 a$ long. Standing at the very top, both distances are equal, and since they add to $2 a$ each is $a$ ; drop a line down and I've made a right triangle, so $b^{2} + c^{2} = a^{2}$ . To describe the shape not the size, I divide: $e = c / a$ . Now I put the Sun on one tack. My closest approach is $a (1 - e)$ , my farthest is $a (1 + e)$ — average them and I recover $a$ , subtract-over-sum and I recover $e$ . If the tacks merge ( $e = 0$ ) it's a circle; if they fly apart ( $e \to 1$ ) it stretches into a parabola and beyond into a hyperbola. One number, $e$ , runs the whole show.

Recall Quick self-test

Why is the constant sum exactly $2 a$ ? ::: At the right vertex the two distances $(a - c)$ and $(a + c)$ add to $2 a$ , and the sum is the same everywhere. Where does $b^{2} = a^{2} (1 - e^{2})$ come from? ::: From the right triangle $b^{2} + c^{2} = a^{2}$ with $c = a e$ . What does $e = 0$ give? ::: A circle: foci merge at the centre, $r_{m i n} = r_{m a x} = a$ . What happens as $e \to 1$ ? ::: The ellipse thins ( $b \to 0$ ) and opens into a parabola/hyperbola.

Connections

Parent: 3.4.06 Kepler's connection — orbits are ellipses (motivation) (Hinglish)
Ellipse — standard equation and focus
Eccentricity of a conic
Focus–directrix definition of conics
Conic sections as slices of a cone
Parabola — projectile motion connection
Hyperbola — unbound orbits and escape
Angular momentum conservation