Sum of focal radii = 2a property: Worked examples

The scenario matrix

Before working anything, let us list every kind of situation the focal-radii property can be asked about. Each later example is tagged with the cell it fills.

#	Case class	What is tricky about it	Example
A	Point with $x > 0$ (right half)	Near the right focus → smaller radius	Ex 1
B	Point with $x < 0$ (left half)	Sign of $e x$ flips; near left focus	Ex 2
C	Point at $x = 0$ (top/bottom co-vertex)	Degenerate: both radii equal	Ex 3
D	Point at a vertex $x = \pm a$	Limiting case: radii are extremes $a (1 \mp e)$	Ex 4
E	Tall ellipse (major axis vertical)	Foci on the $y$ -axis; use $y$ not $x$	Ex 5
F	One radius given, find the other	Use the sum shortcut, no coordinates	Ex 6
G	Build the ellipse from foci + sum	Reverse the property	Ex 7
H	Word problem (whispering gallery)	Translate physical distances → $2 a$	Ex 8
I	Exam twist (locus / two conditions)	Combine sum with another equation	Ex 9

Our running wide ellipse for cases A–D and F is $\frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1,$ so $a^{2} = 25 \Rightarrow a = 5$ , $b^{2} = 9$ , and $e = 1 - \frac{9}{25} = \frac{4}{5}$ . Keep these numbers handy.

Figure — Sum of focal radii = 2a property

The figure above shows this ellipse with its two foci and four sample points, one per cell A–D. Watch how the two coloured focal radii trade length as the point slides.

Case A — point in the right half

Case B — point in the left half

Case C — point at $x = 0$ (degenerate: equal radii)

Worked example Ex 3 (Cell C): the co-vertex

Same ellipse, $P = (0, 3)$ (top of the oval). Find both radii.

Forecast: $P$ sits dead centre horizontally — equally far from both foci. So the two radii should be equal, each half of $2 a$ , i.e. $5$ .

Step 1 — apply the formulas at $x = 0$ . $S P = a - e \cdot 0 = a = 5$ and $S^{'} P = a + e \cdot 0 = a = 5$ . Why this step? At $x = 0$ the $e x$ term vanishes, collapsing both formulas to $a$ . This is the degenerate symmetric case.

Step 2 — geometric cross-check by triangle. $P = (0, 3)$ , foci at $(\pm 4, 0)$ (since $a e = 5 \cdot \frac{4}{5} = 4$ ). Distance $= 4^{2} + 3^{2} = 25 = 5$ . Why this step? Confirms the algebra with the raw distance formula — a right triangle of legs $4$ and $3$ gives hypotenuse $5$ .

Verify: sum $= 5 + 5 = 10 = 2 a$ ✓, and both radii equal as forecast. Note this is also where each focal radius equals $a$ — a handy landmark.

Case D — point at a vertex (limiting extremes)

Worked example Ex 4 (Cell D): the endpoints of the major axis

Same ellipse. Evaluate the radii at the two vertices $(\pm 5, 0)$ .

Forecast: The right vertex $(5, 0)$ is the closest any point gets to $S$ , so $S P$ should hit its minimum; $S^{'} P$ its maximum. These are the extreme values of a focal radius.

Step 1 — right vertex $(a, 0) = (5, 0)$ . $S P = a - e a = a (1 - e) = 5 (1 - \frac{4}{5}) = 1$ ; $S^{'} P = a (1 + e) = 5 (\frac{9}{5}) = 9$ . Why this step? Setting $x = a$ gives the algebraic extremes $a (1 - e)$ and $a (1 + e)$ .

Step 2 — left vertex $(- a, 0) = (- 5, 0)$ . By symmetry $S P = a (1 + e) = 9$ , $S^{'} P = a (1 - e) = 1$ . Why this step? Mirror of Step 1; confirms the smallest possible focal radius is $a (1 - e)$ and the largest is $a (1 + e)$ .

Verify: each vertex gives $1 + 9 = 10 = 2 a$ ✓. Range of any single focal radius: from $a (1 - e) = 1$ up to $a (1 + e) = 9$ . This kills the " $2 b$ " myth — even at the extreme vertex the sum is $2 a = 10$ , never $2 b = 6$ .

Case E — tall ellipse (major axis vertical)

Worked example Ex 5 (Cell E): foci on the

$y$ -axis Ellipse $\frac{x ^{2}}{9} + \frac{y ^{2}}{25} = 1$ , point $P = (9/5, 4)$ . Find the focal radii.

Forecast: The bigger denominator ( $25$ ) sits under $y^{2}$ , so the major axis is vertical. The foci lie on the $y$ -axis and the working variable is $y$ , not $x$ . Sum still $= 2 a$ where $a^{2} = 25$ .

Step 1 — identify the long axis. Since $25 > 9$ , $a^{2} = 25 \Rightarrow a = 5$ (vertical), $b^{2} = 9$ . Why this step? The label " $a$ " always attaches to the larger denominator; foci and the $\pm e \cdot (coordinate)$ rule follow the major axis.

Step 2 — eccentricity. $e = 1 - \frac{b ^{2}}{a ^{2}} = 1 - \frac{9}{25} = \frac{4}{5}$ . Foci at $(0, \pm a e) = (0, \pm 4)$ . Why this step? Same $b^{2} = a^{2} (1 - e^{2})$ relation; only the axis changed.

Step 3 — radii using $y$ . Upper focus $S (0, 4)$ : $S P = a - ey = 5 - \frac{4}{5} (4) = \frac{9}{5}$ . Lower focus $S^{'} (0, - 4)$ : $S^{'} P = a + ey = 5 + \frac{16}{5} = \frac{41}{5}$ . Why this step? Rotate the whole picture $9 0^{\circ}$ : the $x$ in $a \pm e x$ becomes $y$ because the foci moved to the $y$ -axis.

Verify: sum $= \frac{9}{5} + \frac{41}{5} = 10 = 2 a$ ✓. Notice this is Ex 1 stood on its end — same numbers, axes swapped.

Case F — one radius given, find the other

Case G — build the ellipse from the property

Case H — real-world word problem

Worked example Ex 8 (Cell H): the whispering gallery

An elliptical hall has its two "focus" spots $20$ m apart. A whisper at one focus is heard clearly at the other; the sound travels along a wall path whose two segments (whisperer→wall→listener) always total $52$ m. Find the ellipse describing the wall (foci on the $x$ -axis).

Forecast: The two sound segments are exactly the two focal radii of the reflection point — so their fixed total $52$ is $2 a$ . Foci $20$ m apart means $2 a e = 20$ .

Step 1 — translate the physics. For a point on the wall, (distance to one focus) + (distance to other focus) $= 52$ . That constant sum is the ellipse's $2 a$ , so $a = 26$ . Why this step? The reflection property is a physical realisation of "sum of focal radii $= 2 a$ ."

Step 2 — use the focal separation. Foci $20$ m apart $\Rightarrow 2 a e = 20 \Rightarrow a e = 10 \Rightarrow e = \frac{10}{26} = \frac{5}{13}$ . Why this step? Distance between foci is $2 a e$ .

Step 3 — find $b^{2}$ . $b^{2} = a^{2} - a^{2} e^{2} = a^{2} - (a e)^{2} = 2 6^{2} - 1 0^{2} = 676 - 100 = 576$ , so $b = 24$ . Why this step? $c = a e$ gives $b^{2} = a^{2} - c^{2}$ directly, dodging fractions.

Verify: $\frac{x ^{2}}{676} + \frac{y ^{2}}{576} = 1$ . Units: all lengths in metres, consistent ✓. Half-major axis $a = 26$ m so full length $52$ m matches the string total ✓; foci at $\pm 10$ m give the stated $20$ m gap ✓.

Case I — exam-style twist (two conditions)

Worked example Ex 9 (Cell I): find the point, not just the radii

On $\frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1$ , find the point(s) $P$ in the upper half where $S^{'} P = 2 S P$ (the left-focus radius is twice the right-focus radius).

Forecast: Two facts fight it out: $S P + S^{'} P = 10$ (sum) and $S^{'} P = 2 S P$ (given). Solve them together — a point far from $S$ , hence in the right half ( $x > 0$ )? Let us see.

Step 1 — combine sum with the ratio. Substitute $S^{'} P = 2 S P$ into $S P + S^{'} P = 10$ : $3 S P = 10 \Rightarrow S P = \frac{10}{3}$ , $S^{'} P = \frac{20}{3}$ . Why this step? The constant-sum property turns a two-unknown problem into one equation.

Step 2 — solve for $x$ . $S P = a - e x \Rightarrow \frac{10}{3} = 5 - \frac{4}{5} x \Rightarrow \frac{4}{5} x = 5 - \frac{10}{3} = \frac{5}{3} \Rightarrow x = \frac{25}{12}$ . Why this step? Turn the radius back into a coordinate. Positive $x$ ⇒ right half, so $P$ is actually closer to $S$ — consistent with $S P$ being the smaller radius.

Step 3 — solve for $y$ (upper half). $\frac{y ^{2}}{9} = 1 - \frac{x ^{2}}{25} = 1 - \frac{( 25/12 ) ^{2}}{25} = 1 - \frac{25}{144} = \frac{119}{144}$ . So $y^{2} = \frac{119 \cdot 9}{144} = \frac{119}{16}$ , giving $y = \frac{119}{4}$ (positive for upper half). Why this step? The point must also lie on the ellipse; that pins $y$ .

Verify: $S P = 5 - \frac{4}{5} \cdot \frac{25}{12} = 5 - \frac{20}{12} = 5 - \frac{5}{3} = \frac{10}{3}$ ✓ and $S^{'} P = 5 + \frac{5}{3} = \frac{20}{3} = 2 \cdot \frac{10}{3}$ ✓. Sum $\frac{10}{3} + \frac{20}{3} = 10 = 2 a$ ✓. The point is $P = (\frac{25}{12}, \frac{119}{4})$ .

Active recall

Recall Which variable goes into

$a \pm e (\cdot)$ for a tall ellipse, and why? $y$ — because for a tall ellipse the major axis is vertical, the foci sit on the $y$ -axis, so the focal radius depends on the vertical coordinate.

Recall What are the smallest and largest possible focal radii, and where do they occur?

Smallest $= a (1 - e)$ and largest $= a (1 + e)$ , both at the vertices $(\pm a, 0)$ (endpoints of the major axis).

Recall At

$x = 0$ what is special about the two focal radii? They are equal, each exactly $a$ (half of $2 a$ ), since the $e x$ term vanishes.

Connections

Sum of focal radii = 2a property (index 3.4.5) — the parent property these cases exercise.
Ellipse - Standard Equation — where $a, b, e$ and the wide/tall distinction come from.
Eccentricity of Conics — the $e$ inside every $a \pm e (\cdot)$ .
Directrix and Focal Distance — the alternative route to focal distance.
Hyperbola - Difference of Focal Radii = 2a — the difference-cousin for contrast.
Definition of Conic as Locus — the constant-sum locus that underlies Ex 7 and Ex 8.