Difference of focal radii = 2a property: Visual walkthrough

Step 1 — Draw the stage: two foci and a moving point

WHAT. We place two fixed points on a horizontal line, one to the right, one to the left of the centre. Call the right one $S$ and the left one $S^{'}$ (read " $S$ -prime"). A third point $P$ is free to roam. The two straight segments from $P$ down to each focus are the focal radii: $r_{1} = S P$ (distance from $P$ to the right focus) and $r_{2} = S^{'} P$ (distance to the left focus).

WHY. A hyperbola is born from two focus points — you cannot draw one with a single focus. So before any formula, we need the picture of "a point and its two leashes to the two foci."

PICTURE. In the figure, $S$ and $S^{'}$ sit on the horizontal axis. The cyan segment is $r_{1}$ , the amber segment is $r_{2}$ . Notice: as $P$ slides around, both leashes stretch and shrink — the whole trick will be that a certain combination of them stays frozen.

Figure — Difference of focal radii = 2a property

Step 2 — Write each leash length with Pythagoras

WHAT. The length of a segment between two points is just the hypotenuse of a right triangle whose legs are the horizontal gap and the vertical gap. Drop a vertical from $P$ to the axis: the horizontal leg from $S$ is $(x - a e)$ , the vertical leg is $y$ . So

$r_{1} = hypotenuse = S P (x - a e) horizontal gap to S^{2} + y height of P^{2}, r_{2} = (x + a e)^{2} + y^{2} .$

WHY use Pythagoras and not something fancier? Distance in the plane is the hypotenuse of a right triangle — that is the most direct tool that turns "length of a slanted segment" into ordinary $x$ and $y$ . Nothing about the hyperbola is assumed yet; this holds for any point $P$ .

PICTURE. Two shaded right triangles share the same vertical leg $y$ (the height of $P$ ). Their horizontal legs differ only by the sign in front of $a e$ : $x - a e$ reaches the right focus, $x + a e$ reaches the left one.

Step 3 — Feed in the curve to kill the $y$

WHAT. So far $r_{1}, r_{2}$ still contain $y$ , which changes from point to point. We remove $y$ using the fact that $P$ lives on the hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ . Solve it for $y^{2}$ :

$y^{2} = b^{2} (\frac{x ^{2}}{a ^{2}} - 1) .$

Then use the hyperbola's defining link between $b$ and $e$ , namely $b^2=a^2(e^2-1)$:

$y^{2} = this is b^{2} a^{2} (e^{2} - 1) (\frac{x ^{2}}{a ^{2}} - 1) .$

WHY this substitution? $r_{2}$ mixes $x$ and $y$ . If we can express $y^{2}$ purely through $x$ , then $r_{2}$ becomes a function of $x$ alone — and a function of one variable is something we can hope to simplify into a perfect square. Choosing $b^{2} = a^{2} (e^{2} - 1)$ (not the ellipse version $a^{2} (1 - e^{2})$ ) is what makes the algebra collapse; with $e > 1$ this keeps $b^{2} > 0$ .

PICTURE. The figure shows the curve; at a chosen $x$ , the height $y$ of the point is fixed by the curve. The dashed vertical shows $y$ being "read off" the curve rather than left free.

Step 4 — Watch the mess collapse into a perfect square

WHAT. Substitute that $y^{2}$ into $r_{2}^{2} = (x + a e)^{2} + y^{2}$ and expand:

$r_{2}^{2} = x^{2} + 2 a e x + a^{2} e^{2} (x + a e)^{2} + from y^{2} (e^{2} - 1) x^{2} - a^{2} (e^{2} - 1) .$

Gather like terms:

$r_{2}^{2} = x^{2} terms e^{2} x^{2} + cross term 2 a e x + constant a^{2} = (e x + a)^{2} .$

So taking the square root, $r_{2} = ∣ e x + a ∣$ .

WHY does this matter? A square root of a perfect square is just the thing inside (up to sign). All the ugly geometry has boiled down to the tidy linear expression $e x + a$ . This only worked because we plugged in the correct $b$ – $e$ relation in Step 3.

PICTURE. The figure lines up the three surviving pieces $e^{2} x^{2}$ , $2 a e x$ , $a^{2}$ as the three tiles of the algebraic identity $(e x + a)^{2} = e^{2} x^{2} + 2 a e x + a^{2}$ — a literal "completing the square" picture.

Step 5 — Do the twin computation for $r_{1}$

WHAT. Repeat Step 4 but with the right focus, i.e. $(x - a e)$ instead of $(x + a e)$ . The only change is a sign in the cross term:

$r_{1}^{2} = e^{2} x^{2} - 2 a e x + a^{2} = (e x - a)^{2} \Rightarrow r_{1} = ∣ e x - a ∣.$

WHY. The two foci sit symmetrically at $\pm a e$ , so the computation is a mirror image; only the middle sign flips. Doing it explicitly reassures us there is no hidden asymmetry.

PICTURE. Side-by-side: the $r_{2}$ tiling from Step 4 versus the $r_{1}$ tiling, with the single flipped middle tile highlighted in amber.

Step 6 — Resolve the moduli branch by branch

WHAT. A distance is never negative, so we must decide the sign inside each modulus. This depends on which branch of the hyperbola $P$ is on.

Right branch ( $x \geq a > 0$ ): since $e > 1$ , we have $e x \geq e a > a$ , so $e x - a > 0$ and $e x + a > 0$ . Then $r_{1} = e x - a$ and $r_{2} = e x + a$ .
Left branch ( $x \leq - a < 0$ ): now $e x \leq - e a < - a$ , so $e x + a < 0$ and $e x - a < 0$ . The moduli flip the signs: $r_{1} = - (e x - a) = a - e x$ and $r_{2} = - (e x + a) = - a - e x$ .

WHY cover both? If we only ever looked at the right branch we would miss that the sign of the difference reverses. That reversal is exactly why the definition carries an absolute value $∣ S P - S^{'} P ∣$ .

PICTURE. Both branches are drawn. On the right branch the left focus $S^{'}$ is the farther one (amber leash longer); on the left branch the right focus $S$ is farther. The "farther focus" swaps sides.

Step 7 — Subtract, and watch $x$ vanish

WHAT. On the right branch:

$r_{2} - r_{1} = (e x + a) - (e x - a) = = 0 e x - e x + = 2 a a + a = 2 a .$

On the left branch:

$r_{1} - r_{2} = (a - e x) - (- a - e x) = 2 a .$

Either way the answer is the same fixed number $2 a$ , and each way the term carrying $x$ cancels.

WHY this is the whole point. The $e x$ terms are the only place $x$ entered. When they cancel, the result no longer depends on where $P$ is — it is the same constant for every point on the curve. That constancy is the hyperbola property. Because the sign of the difference flips between branches, we write both cases at once as

$∣ S P - S^{'} P ∣ = 2 a .$

PICTURE. Two positions of $P$ (near the tip and far out) share the same amber "gap" $= 2 a$ , even though each individual leash is wildly different in length.

Step 8 — The degenerate check: the sum does not freeze

WHAT. For contrast, add the radii on the right branch:

$r_{1} + r_{2} = (e x - a) + (e x + a) = 2 e x .$

This grows with $x$ — it is not constant.

WHY show this? It is the fingerprint that separates the two conics. The ellipse fixes the SUM; the hyperbola fixes the difference. If you ever get a constant sum, you drew an ellipse by mistake.

PICTURE. As $P$ marches outward, the amber difference stays a fixed width while the cyan sum stretches without bound — the two behaviours side by side.

The one-picture summary

Everything in one diagram: the two triangles from Step 2, the substitution that erased $y$ , the perfect squares $e x \pm a$ , and the cancellation that leaves the frozen gap $2 a$ regardless of $P$ 's position.

Recall Feynman retelling — the whole walkthrough in plain words

Picture two flagpoles, $S$ and $S^{'}$ , planted on a line. Stand at a point $P$ and stretch two ropes, one to each pole; their lengths are $r_{1}$ and $r_{2}$ . First, we measured each rope with Pythagoras — a rope is the slanted side of a right triangle whose legs are "how far across" and "how high up." That left ropes full of $y$ (the height), which is annoying because $y$ changes as you walk. So we used the curve's own equation to trade $y$ for $x$ , choosing the special ingredient $b^{2} = a^{2} (e^{2} - 1)$ that makes hyperbolas work. When we did the algebra, the horrible square roots melted into neat little expressions: $r_{2} = e x + a$ and $r_{1} = e x - a$ . Then the magic: subtract them. The $e x$ pieces — the only pieces that knew where you were standing — cancelled, leaving just $2 a$ . So no matter where you roam, the two ropes always differ by the same amount, $2 a$ , the gap between the curve's two tips. On the far branch the closer and farther poles swap, which only flips the sign, so we wrap it in absolute-value bars: $∣ S P - S^{'} P ∣ = 2 a$ . And just to be safe we added the ropes instead — that gave $2 e x$ , which keeps growing, proving the difference (not the sum) is the hyperbola's signature.

Connections

Parent: Difference of focal radii = 2a
Hyperbola — Standard Equation and Definition
Ellipse — Sum of Focal Radii = 2a
Eccentricity of Conics
b² = a²(e²−1) relation
Latus Rectum of a Hyperbola
Focus–Directrix Property