Sum of focal radii = 2a property: Exercises — Maths deep dive

This is the practice page for the parent property. Every problem is stated cleanly, then fully solved inside a collapsible Solution box so you can test yourself first. Work the problem, then reveal.

The two facts we lean on the whole way down: $S P = a - e x, S^{'} P = a + e x, S P + S^{'} P = 2 a$ Here $S (a e, 0)$ is the right focus, $S^{'} (- a e, 0)$ the left focus, $a$ is the semi-major axis, and $e$ is the eccentricity (a number between $0$ and $1$ for an ellipse). If any of those symbols feel unfamiliar, re-read the parent note before starting.

Below, " $L 1 \to L 5$ " is a difficulty ladder:

L1 Recognition — spot the constant, read off $a$ .
L2 Application — plug into $a \pm e x$ .
L3 Analysis — reason backwards, combine two facts.
L4 Synthesis — build an ellipse / prove a small result.
L5 Mastery — multi-step, no formula handed to you.

Figure — Sum of focal radii = 2a property

The figure above is our reference picture for the whole page: an ellipse, its two foci $S$ and $S^{'}$ , a moving point $P$ , and the two focal radii $r_{1} = S P$ (cyan) and $r_{2} = S^{'} P$ (amber). Their sum is the taut string — always $2 a$ .

Level 1 — Recognition

Recall Solution 1.1

WHAT we use: the whole property is $S P + S^{'} P = 2 a$ . Nothing else is needed. WHY it works: the sum is a fixed constant for every point — so knowing one radius and the constant pins the other. $S^{'} P = 2 a - S P = 10 - 3 = 7.$ Answer: $S^{'} P = 7$ .

Recall Solution 1.2

WHAT to read off: the larger denominator sits under $x^{2}$ ( $49 > 24$ ), so the major axis is along the $x$ -axis and $a^{2} = 49$ . WHY the larger one: $a > b$ always, and $a^{2}$ is the bigger of the two denominators. $a = 49 = 7 \Rightarrow 2 a = 14.$ Answer: the focal radii of every point add to $14$ .

Level 2 — Application

Recall Solution 2.1

Step 1 — read $a, b$ . $a^{2} = 25 \Rightarrow a = 5$ ; $b^{2} = 16$ . Step 2 — get $e$ . From $b^{2} = a^{2} (1 - e^{2})$ : $e = 1 - \frac{16}{25} = \frac{9}{25} = \frac{3}{5} .$ Why this tool: eccentricity is exactly what sits inside $a \pm e x$ , so we must compute it first. Step 3 — apply the formulas with $x = 3$ : $S P = a - e x = 5 - \frac{3}{5} (3) = 5 - \frac{9}{5} = \frac{16}{5},$ $S^{'} P = a + e x = 5 + \frac{9}{5} = \frac{34}{5} .$ Step 4 — verify. $S P + S^{'} P = \frac{16}{5} + \frac{34}{5} = \frac{50}{5} = 10 = 2 a .$ ✓ Answer: $S P = \frac{16}{5}$ , $S^{'} P = \frac{34}{5}$ .

Recall Solution 2.2

Step 1: $a = 10$ , $b^{2} = 64$ . Step 2: $e = 1 - \frac{64}{100} = \frac{36}{100} = \frac{3}{5} .$ Step 3: with $x = - 5$ , $S P = a - e x = 10 - \frac{3}{5} (- 5) = 10 + 3 = 13,$ $S^{'} P = a + e x = 10 + \frac{3}{5} (- 5) = 10 - 3 = 7.$ Check: $13 + 7 = 20 = 2 a .$ ✓ Answer: $S P = 13$ , $S^{'} P = 7$ . (Notice: negative $x$ means $P$ is on the left, so it is nearer $S^{'}$ — and indeed $S^{'} P = 7$ is the smaller one.)

Level 3 — Analysis

Recall Solution 3.1

Step 1 — forecast $S^{'} P$ first. $a^{2} = 16 \Rightarrow 2 a = 8$ , so before any coordinates: $S^{'} P = 2 a - S P = 8 - 5 = 3.$ Step 2 — now find $x$ . $e = 1 - \frac{7}{16} = \frac{9}{16} = \frac{3}{4} .$ Then $S P = a - e x = 4 - \frac{3}{4} x = 5 \Rightarrow \frac{3}{4} x = - 1 \Rightarrow x = - \frac{4}{3} .$ Step 3 — verify with the formula. $S^{'} P = a + e x = 4 + \frac{3}{4} (- \frac{4}{3}) = 4 - 1 = 3.$ ✓ Matches the forecast. Answer: $x = - \frac{4}{3}$ , $S^{'} P = 3$ .

Recall Solution 3.2

Step 1 — split the constant. Let $S P = 2 k$ , $S^{'} P = 3 k$ . Their sum is $2 a$ : $2 k + 3 k = 20 \Rightarrow 5 k = 20 \Rightarrow k = 4.$ So $S P = 8$ , $S^{'} P = 12$ . Step 2 — extract $e x$ . Recall $S^{'} P - S P = 2 e x$ (subtracting the two formulas): $2 e x = 12 - 8 = 4 \Rightarrow e x = 2.$ WHY this works: the sum fixes the ratio's total, the difference reveals the $e x$ -offset — the two facts together fully locate the point along $x$ . Answer: $S P = 8$ , $S^{'} P = 12$ , $e x = 2$ .

Level 4 — Synthesis

Recall Solution 4.1

Step 1 — $a$ from the sum. $2 a = 10 \Rightarrow a = 5$ , so $a^{2} = 25$ . Step 2 — $a e$ from the foci. Foci at $(\pm a e, 0) = (\pm 4, 0) \Rightarrow a e = 4 \Rightarrow e = \frac{4}{5} .$ Step 3 — $b^{2}$ . Use $b^{2} = a^{2} (1 - e^{2}) = 25 (1 - \frac{16}{25}) = 25 - 16 = 9.$ (Faster: $b^{2} = a^{2} - (a e)^{2} = 25 - 16 = 9$ .) Step 4 — write it. $\frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1.$ Answer: $\frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1$ .

Recall Solution 4.2

Step 1 — use symmetry at $(0, 4)$ . At the top of the minor axis, the point is equidistant from both foci (both are $(\pm 3, 0)$ , symmetric about the $y$ -axis). Each focal radius is $S P = S^{'} P = (0 - 3)^{2} + (4 - 0)^{2} = 9 + 16 = 5.$ Step 2 — apply the property. $2 a = S P + S^{'} P = 5 + 5 = 10$ , so $a = 5$ , $a^{2} = 25$ . WHY this shortcut: at the minor-axis end the two radii are equal, each equal to $a$ itself — a clean geometric anchor (indeed $b^{2} + (a e)^{2} = a^{2}$ is exactly this right triangle: $16 + 9 = 25$ ). Step 3 — $b^{2}$ . The point $(0, 4)$ is a minor-axis end, so $b = 4$ , $b^{2} = 16$ . Step 4 — equation: $\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1.$ Answer: $2 a = 10$ , ellipse $\frac{x ^{2}}{25} + \frac{y ^{2}}{16} = 1$ .

The figure shows Exercise 4.2's right triangle: from the minor-axis top $(0, b)$ down to a focus $(a e, 0)$ , the hypotenuse has length exactly $a$ . That is why each focal radius there equals $a$ , and the two of them add to $2 a$ .

Level 5 — Mastery

Recall Solution 5.1

Step 1 — constants. $a^{2} = 169 \Rightarrow a = 13$ ; $b^{2} = 144$ ; $e = 1 - \frac{144}{169} = \frac{25}{169} = \frac{5}{13} .$ Step 2 — set up the ratio. $S^{'} P = \frac{7}{4} S P$ . Also $S P + S^{'} P = 2 a = 26$ . Substitute: $S P + \frac{7}{4} S P = 26 \Rightarrow \frac{11}{4} S P = 26 \Rightarrow S P = \frac{104}{11} .$ Step 3 — solve for $x$ from $S P = a - e x$ : $\frac{104}{11} = 13 - \frac{5}{13} x \Rightarrow \frac{5}{13} x = 13 - \frac{104}{11} = \frac{143 - 104}{11} = \frac{39}{11} .$ $x = \frac{13}{5} \cdot \frac{39}{11} = \frac{507}{55} = \frac{507}{55} .$ Step 4 — find $y$ from the ellipse equation: $y^{2} = 144 (1 - \frac{x ^{2}}{169}) .$ With $x = \frac{507}{55}$ : $\frac{x ^{2}}{169} = \frac{( 507/55 ) ^{2}}{169} = \frac{( 3 \cdot 169/55 ) ^{2}}{169} = \frac{9 \cdot 169}{5 5 ^{2}} = \frac{1521}{3025} .$ So $y^{2} = 144 \cdot \frac{3025 - 1521}{3025} = 144 \cdot \frac{1504}{3025} = \frac{216576}{3025},$ $y = \pm \frac{216576}{55} = \pm \frac{465.4 \dots}{55} \approx \pm 8.462.$ Answer: two points, $(\frac{507}{55}, \pm \frac{216576}{55}) \approx (9.22, \pm 8.46)$ . Why two points: the equation fixes $x$ but $y$ can be $\pm$ — the ellipse is symmetric about the $x$ -axis, and both mirror points share the same focal radii.

Recall Solution 5.2

Proof. Take $P = (x, y)$ on the ellipse. The distance to the right focus $S (a e, 0)$ is $S P = (x - a e)^{2} + y^{2} .$ From the ellipse equation and $b^{2} = a^{2} (1 - e^{2})$ , $y^{2} = b^{2} (1 - \frac{x ^{2}}{a ^{2}}) = (a^{2} - a^{2} e^{2}) - x^{2} + e^{2} x^{2} .$ Expand the inside of the root: $(x - a e)^{2} + y^{2} = x^{2} - 2 a e x + a^{2} e^{2} + a^{2} - a^{2} e^{2} - x^{2} + e^{2} x^{2} = a^{2} - 2 a e x + e^{2} x^{2} = (a - e x)^{2} .$ Since $- a \leq x \leq a$ and $0 < e < 1$ , $a - e x > 0$ , so $S P = a - e x .$ Identically for $S^{'} (- a e, 0)$ : $S^{'} P = a + e x$ . Adding, $S P + S^{'} P = (a - e x) + (a + e x) = 2 a . ■$ Numerical part. $S P = a - e x = 6 - \frac{1}{3} (3) = 6 - 1 = 5.$ Answer: $S P = 5$ (and $S^{'} P = 7$ , sum $= 12 = 2 a$ ✓).

Wrap-up recall

Recall Which fact solved which level?

L1 recognition — read $2 a$ off the constant ::: The sum $S P + S^{'} P = 2 a$ and $a^{2}$ = larger denominator. L2 application — get numbers from $a \pm e x$ ::: Compute $e$ first, then plug $x$ . L3 analysis — reason backwards ::: Sum fixes total; difference $S^{'} P - S P = 2 e x$ (keep the sign). L4 synthesis — build the ellipse ::: Use raw distance formula or minor-axis symmetry to get $a$ before touching $a \pm e x$ . L5 mastery — prove + apply ::: $(x - a e)^{2} + y^{2} = (a - e x)^{2}$ ; justify $a - e x > 0$ before rooting.

Connections

Sum of Focal Radii = 2a property — the parent property these drill.
Ellipse - Standard Equation — where $a, b, e$ come from.
Eccentricity of Conics — the $e$ inside $a \pm e x$ .
Directrix and Focal Distance — an alternate route to focal radii.
Hyperbola - Difference of Focal Radii = 2a — the sign-flip cousin (L5 trap).
Definition of Conic as Locus — ellipse as constant-sum locus.