Sum of focal radii = 2a property: Visual walkthrough

What this page is: the parent result $S P + S^{'} P = 2 a$ , rebuilt from a blank picture, one drawing per step. Nothing is assumed — every symbol is drawn before it is used. Follow the coloured arrows in each figure; the words only describe what you can already see.

Step 0 — What are we even looking at?

Before a single formula, let us just meet the players in a drawing.

Figure — Sum of focal radii = 2a property

Everything below is just: find $r_{1}$ , find $r_{2}$ , add them.

Step 1 — Pin down where the foci sit

WHAT. We place the ellipse on graph paper so its centre is the origin, and we write down the exact coordinates of the two foci.

WHY. To measure a distance we need coordinates. The moment we have numbers for $S$ and $S^{'}$ , the distance formula can do its job.

Let us decode $a e$ term by term:

$a$ = the half-length of the major axis (centre to tip).
$e$ = the eccentricity, a number between $0$ and $1$ measuring "how squashed" the oval is (see Eccentricity of Conics). $e = 0$ is a perfect circle; $e$ near $1$ is very flat.
So $a e$ = a fraction $e$ of the half-width $a$ — a point that sits inside the oval, part-way out towards the tip. The right focus is at $+ a e$ , the left one at the mirror position $- a e$ .

Step 2 — Write one focal radius with the distance formula

WHAT. Measure the straight-line length from $P = (x, y)$ to the right focus $S = (a e, 0)$ .

WHY this tool — the distance formula, not anything else? A focal radius is a straight-line length between two points. The one tool that turns two pairs of coordinates into a length is the Pythagorean distance formula: horizontal gap and vertical gap are the two legs of a right triangle, and the radius is its hypotenuse.

Right now $r_{1}$ depends on both $x$ and $y$ . That is messy. The next step removes $y$ .

Step 3 — Trade $y^{2}$ away using the ellipse equation

WHAT. Replace $y^{2}$ with an expression in $x$ only.

WHY. $P$ is not any old point — it lives on the ellipse. That constraint is a free equation linking $y$ to $x$ . Using it lets us describe the radius with a single variable $x$ , which is what makes the final sum collapse.

Solve the ellipse equation for $y^{2}$ : $y^{2} = b^{2} (1 - \frac{x ^{2}}{a ^{2}}) .$

Now bring in the one relationship that ties $b$ , $a$ , $e$ together (from Eccentricity of Conics): $b^{2} = a^{2} (1 - e^{2}) = a^{2} - a^{2} e^{2} .$

Substitute and expand: $y^{2} = (a^{2} - a^{2} e^{2}) (1 - \frac{x ^{2}}{a ^{2}}) = a^{2} - a^{2} e^{2} - x^{2} + e^{2} x^{2} .$

Term by term, this says: the height² at position $x$ is built from four pieces — two constants ( $a^{2}$ , $- a^{2} e^{2}$ ) and two $x$ -dependent pieces ( $- x^{2}$ , $+ e^{2} x^{2}$ ). Watch those last two: they are exactly the terms that will cancel the mess in Step 4.

Step 4 — Watch the square root collapse to a perfect square

WHAT. Put the new $y^{2}$ back inside $r_{1}$ , expand, and cancel.

WHY. This is the payoff of using $b^{2} = a^{2} (1 - e^{2})$ : the ugly root turns into something whose square root is clean and linear in $x$ .

Expand the inside of the root: $(x - a e)^{2} + y^{2} = from (x - a e)^{2} x^{2} - 2 a e x + a^{2} e^{2} + from y^{2} a^{2} - a^{2} e^{2} - x^{2} + e^{2} x^{2} .$

Now cancel colour-by-colour:

$x^{2}$ and $- x^{2}$ cancel.
$a^{2} e^{2}$ and $- a^{2} e^{2}$ cancel.

What survives: $= a^{2} - 2 a e x + e^{2} x^{2} = (a - e x)^{2} .$

That last equality is just the algebra identity $(a - e x)^{2} = a^{2} - 2 a e x + e^{2} x^{2}$ read backwards.

Step 5 — Take the root (and check the sign carefully)

WHAT. $(a - e x)^{2} = ∣ a - e x ∣$ . We need to know it is $a - e x$ , not $- (a - e x)$ .

WHY this matters — the sign case. A square root is always the non-negative value. So we must confirm $a - e x$ itself is already positive; otherwise we'd need to flip its sign.

Check the extremes of $e x$ :

Largest $e x$ happens at $x = a$ : then $e x = e a < a$ (since $e < 1$ ). So $a - e x > 0$ . ✓
Smallest $e x$ happens at $x = - a$ : then $e x = - e a$ , so $a - e x = a + e a > 0$ . ✓

Across the whole allowed range, $a - e x$ stays positive, so the root is simply: $r_{1} = S P = a - e x$

Step 6 — The other focus, by mirror symmetry

WHAT. Repeat everything for the left focus $S^{'} = (- a e, 0)$ .

WHY we don't redo the algebra. The only change from Step 2 is $a e \to - a e$ , i.e. $(x - a e)$ becomes $(x + a e)$ . Every cancellation happens identically, but the surviving square is $(a + e x)^{2}$ .

$r_{2} = S^{'} P = (x + a e)^{2} + y^{2} = (a + e x)^{2} = a + e x$ $r_{2} = S^{'} P = a + e x$

(Same positivity check: $a + e x \geq a - e a > 0$ , always positive.)

Step 7 — Add, and watch $x$ vanish

WHAT. Sum the two boxed results.

WHY this is the whole point. The two radii each carry a $\pm e x$ term. When we add, those terms are equal in size and opposite in sign — they annihilate, leaving a constant with no $x$ in it. A quantity with no $x$ cannot change as $P$ slides around the ellipse.

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

$- e x$ from the near focus and $+ e x$ from the far focus cancel.
What is left, $a + a = 2 a$ , is exactly the length of the major axis — a fixed number.

Step 8 — The degenerate cases (never leave a gap)

WHAT. Test the formula at the corners and at the boundary $e \to 0$ .

WHY. A trustworthy formula must survive its extreme inputs. If it broke at a vertex or at "circle", we'd have made a mistake.

Case	$x$	$r_{1} = a - e x$	$r_{2} = a + e x$	Sum
Right tip (vertex)	$x = a$	$a - e a = a (1 - e)$	$a + e a = a (1 + e)$	$2 a$ ✓
Left tip (vertex)	$x = - a$	$a + e a = a (1 + e)$	$a - e a = a (1 - e)$	$2 a$ ✓
Top of oval	$x = 0$	$a$	$a$	$2 a$ ✓
Circle limit	any $x$ , $e \to 0$	$a$	$a$	$2 a$ ✓

At a vertex, both radii lie flat along the major axis; their lengths are just "half-width minus focus-offset" and "half-width plus focus-offset" — they visibly add to the full width $2 a$ .
At the top ( $x = 0$ ), symmetry forces $r_{1} = r_{2}$ , and $2 r_{1} = 2 a$ gives each equal to $a$ . (This is the picture where each focal radius equals $a$ exactly — a handy landmark.)
As $e \to 0$ the foci both slide to the centre: the ellipse becomes a circle of radius $a$ , and both "focal radii" become the radius $a$ . The property degenerates into "radius + radius = diameter = $2 a$ ", perfectly consistent.

The one-picture summary

Here is the entire derivation compressed into a single diagram: measure $r_{1}$ and $r_{2}$ from $P$ , expose the $\mp e x$ they carry, cancel, and read off $2 a$ .

Recall Feynman retelling — say it to a friend with no maths

Two nails in a board hold a loop of string; a pencil pulls it tight and draws an oval. The string never gets longer, so pencil-to-nail-A plus pencil-to-nail-B is always the same length. To prove it with numbers, we put the oval on graph paper, wrote each of those two distances with Pythagoras, and used the oval's own equation to swap out the height. The algebra magically folded into two neat lengths: one focus gives $a$ minus a wobble $e x$ , the other gives $a$ plus the same wobble $e x$ . Add them — the wobbles kill each other, leaving plain $a + a = 2 a$ , the width of the oval the long way. It works at the tips, at the top, and even when the oval relaxes into a circle. That constancy — sum $= 2 a$ — is what "ellipse" means.

Connections

Parent topic — the full result and worked examples.
Ellipse - Standard Equation — where $a$ , $b$ , $e$ live.
Eccentricity of Conics — the $e$ inside $a \pm e x$ .
Directrix and Focal Distance — focal radius as $e \times$ (distance to directrix).
Hyperbola - Difference of Focal Radii = 2a — the difference cousin.
Definition of Conic as Locus — ellipse as the constant-sum locus.

In Step 2, a focal radius is written as which kind of length?

The hypotenuse of a right triangle (via the distance formula).

Which substitution folds the root into a perfect square?

b^{2} = a^{2} (1 - e^{2})

After expanding,

(x - a e)^{2} + y^{2}

equals?

(a - e x)^{2}

Why do we keep

a - e x

(not its negative) after the square root?

Because

a - e x > 0

for all

- a \leq x \leq a

0 < e < 1

At the top of the ellipse (

x = 0

), each focal radius equals?

a

e \to 0

, the sum-of-focal-radii property becomes?

Circle: radius + radius = diameter =

2 a