Difference of focal radii = 2a property: Worked examples

Before anything: every symbol we use is already earned by the parent. To be safe, in plain words:

$a$ = half the distance between the two tips (vertices) of the hyperbola. Picture the two tips on the $x$ -axis at $x = a$ and $x = - a$ ; $a$ is how far each tip sits from the centre.
$e$ = eccentricity, a number bigger than $1$ that says "how flared open" the hyperbola is.
$a e$ = the $x$ -distance from centre to a focus. A focus is one of the two special fixed points; $S$ sits at $(a e, 0)$ (right), $S^{'}$ at $(- a e, 0)$ (left).
$∣ \cdot ∣$ = "throw away the minus sign" — a distance can never be negative.

The scenario matrix

Every problem this topic can throw is one (or a blend) of these cells. Each example below is stamped with its cell.

Cell	What makes it distinct	Covered by
A. Right branch, ordinary point	$x \geq a$ , both $e x \pm a > 0$ , no modulus needed	Ex 1
B. Left branch, sign flip	$x \leq - a$ , $e x - a$ and $e x + a$ can go negative; modulus bites	Ex 2
C. Vertex (boundary / degenerate)	$x = a$ exactly; one radius hits its smallest value	Ex 3
D. Modulus trap point	a point where $e x - a < 0$ ; must NOT drop the sign	Ex 4
E. Build the curve from the property	given foci + given difference, recover $a, b, e$	Ex 5
F. Limiting / consistency check	is the SUM constant? behaviour as $x \to \infty$	Ex 6
G. Real-world word problem	sound / distance-difference locates a point	Ex 7
H. Exam twist	difference given as a fraction, back out full equation and a radius	Ex 8

Example 1 — Cell A: right branch, ordinary point

Worked example Both focal radii at a clean right-branch point

For $\frac{x ^{2}}{16} - \frac{y ^{2}}{9} = 1$ , find $r_{1}$ and $r_{2}$ at $x = 8$ (right branch), and confirm the difference.

Forecast: Before computing — guess whether $r_{2}$ (distance to the left focus) is bigger or smaller than $r_{1}$ . On the right branch the left focus is farther, so bet $r_{2} > r_{1}$ .

Step 1. Read off $a^{2} = 16 \Rightarrow a = 4$ , $b^{2} = 9$ . Why this step? Standard form hands you $a$ and $b$ for free — no work needed.

Step 2. Find $e$ from $b^{2} = a^{2} (e^{2} - 1)$ : $9 = 16 (e^{2} - 1) \Rightarrow e^{2} - 1 = \frac{9}{16} \Rightarrow e^{2} = \frac{25}{16} \Rightarrow e = \frac{5}{4}$ . Why this step? The radius formulas $∣ e x \pm a ∣$ need $e$ ; this is the only bridge from $b$ to $e$ .

Step 3. $r_{1} = e x - a = \frac{5}{4} (8) - 4 = 10 - 4 = 6$ . $r_{2} = e x + a = 10 + 4 = 14$ . Why this step? $x = 8 > a = 4$ , so $e x > a$ and both quantities are positive — no modulus needed (this is what makes it Cell A).

Verify: $r_{2} - r_{1} = 14 - 6 = 8 = 2 a$ ✓ (since $2 a = 8$ ). And our forecast held: $r_{2} = 14 > r_{1} = 6$ . ✓

Example 2 — Cell B: left branch, sign flip

Worked example The same curve, now on the LEFT branch

Same hyperbola $\frac{x ^{2}}{16} - \frac{y ^{2}}{9} = 1$ , $e = \frac{5}{4}$ , $a = 4$ . Find $r_{1}, r_{2}$ at $x = - 8$ (left branch) and confirm the difference is still $2 a$ but the sign of $S P - S^{'} P$ flips.

Forecast: On the left branch the right focus $S$ is now the far one, so guess $r_{1} = S P > r_{2} = S^{'} P$ — the reverse of Example 1.

Step 1. Compute the raw insides: $e x = \frac{5}{4} (- 8) = - 10$ . So $e x - a = - 10 - 4 = - 14$ and $e x + a = - 10 + 4 = - 6$ . Why this step? Both came out negative — this is exactly why the formulas carry an absolute value. A distance cannot be $- 14$ .

Step 2. Apply the modulus: $r_{1} = ∣ e x - a ∣ = ∣ - 14∣ = 14$ , $r_{2} = ∣ e x + a ∣ = ∣ - 6∣ = 6$ . Why this step? Focal radii are lengths $\geq 0$ ; the modulus repairs the sign the algebra produced.

Step 3. Difference: here $S P - S^{'} P = r_{1} - r_{2} = 14 - 6 = 8 = 2 a$ , so $S P > S^{'} P$ . Why this step? On the left branch the sign flips versus the right branch — hence $∣ S P - S^{'} P ∣ = 2 a$ is the honest statement covering both.

Verify: $∣ r_{1} - r_{2} ∣ = ∣14 - 6∣ = 8 = 2 a$ ✓. Forecast held: $r_{1} > r_{2}$ on the left branch. ✓

Figure — Difference of focal radii = 2a property

Example 3 — Cell C: the vertex (boundary case)

Worked example Radii AT a vertex — the smallest a radius can be

Same curve ( $a = 4$ , $e = \frac{5}{4}$ , foci at $(\pm 5, 0)$ since $a e = \frac{5}{4} \cdot 4 = 5$ ). Find $r_{1}, r_{2}$ at the right vertex $(a, 0) = (4, 0)$ .

Forecast: At the right vertex the point sits between centre and right focus... actually just to the left of $S (5, 0)$ . Guess $r_{1} = S P$ is small (about $1$ ) and $r_{2} = S^{'} P$ is large.

Step 1. At $x = a = 4$ : $r_{1} = e x - a = \frac{5}{4} (4) - 4 = 5 - 4 = 1$ . Why this step? $x = a$ is the smallest $x$ on the right branch, so $r_{1} = e x - a$ is at its minimum here — the vertex is the degenerate/boundary case of Cell A.

Step 2. $r_{2} = e x + a = 5 + 4 = 9$ . Why this step? Same formula, still positive; nothing degenerates for $r_{2}$ .

Step 3. Geometric sanity: distance from vertex $(4, 0)$ to right focus $(5, 0)$ is $5 - 4 = 1 = r_{1}$ ; to left focus $(- 5, 0)$ is $4 - (- 5) = 9 = r_{2}$ . Both match by pure ruler distance on the $x$ -axis. Why this step? At the vertex everything lies on one line, so we can double-check with plain subtraction — a foolproof cross-check that the formula is right at the boundary.

Verify: $r_{2} - r_{1} = 9 - 1 = 8 = 2 a$ ✓. Also $r_{1} = a (e - 1) = 4 (\frac{5}{4} - 1) = 1$ ✓ (the neat vertex identity: nearest-focus distance $= a (e - 1)$ ).

Example 4 — Cell D: the modulus trap (right branch is not enough)

Worked example A point where dropping the modulus gives a wrong, negative "distance"

Take $\frac{x ^{2}}{4} - \frac{y ^{2}}{5} = 1$ . Find $r_{1} = S P$ at the left vertex $x = - 2$ , and show why writing $r_{1} = e x - a$ without the modulus is wrong here.

Forecast: $x = - 2$ is negative, so $e x - a$ will likely be negative — a red flag that the bare formula lies.

Step 1. $a^{2} = 4 \Rightarrow a = 2$ , $b^{2} = 5$ . Then $b^{2} = a^{2} (e^{2} - 1) \Rightarrow 5 = 4 (e^{2} - 1) \Rightarrow e^{2} = \frac{9}{4} \Rightarrow e = \frac{3}{2}$ . Why this step? We need $e$ before any radius; standard bridge.

Step 2. Naive (wrong): $e x - a = \frac{3}{2} (- 2) - 2 = - 3 - 2 = - 5$ . A distance of $- 5$ is nonsense. Why this step? This exposes the trap — the algebra of " $e x - a$ " is only the right-branch shortcut; the honest formula is $∣ e x - a ∣$ .

Step 3. Correct: $r_{1} = ∣ e x - a ∣ = ∣ - 5∣ = 5$ . Why this step? The modulus is not decoration — on the left branch it is doing real work.

Verify: Ruler check: left vertex is $(- 2, 0)$ , right focus $S = (a e, 0) = (\frac{3}{2} \cdot 2, 0) = (3, 0)$ . Distance $= 3 - (- 2) = 5 = r_{1}$ ✓. And $r_{2} = ∣ e x + a ∣ = ∣ - 3 + 2∣ = ∣ - 1∣ = 1$ , so $∣ r_{1} - r_{2} ∣ = ∣5 - 1∣ = 4 = 2 a$ ✓.

Example 5 — Cell E: build the curve from the property

Worked example Given foci and the constant difference, recover the equation

A point moves so its distances from $(\pm 13, 0)$ always differ by $10$ . Find its equation.

Forecast: The two fixed points are foci; "differ by a constant" screams hyperbola with $2 a = 10$ . Guess $a = 5$ .

Step 1. The constant difference is $2 a$ by definition, so $2 a = 10 \Rightarrow a = 5$ . Why this step? The defining property gives $a$ instantly — no algebra.

Step 2. Foci at $(\pm 13, 0)$ means $a e = 13$ . With $a = 5$ : $e = \frac{13}{5}$ . Why this step? The focus $x$ -coordinate is $a e$ ; knowing $a$ turns it into $e$ .

Step 3. $b^{2} = a^{2} (e^{2} - 1) = 25 (\frac{169}{25} - 1) = 169 - 25 = 144$ . Why this step? The hyperbola $b$ – $e$ relation converts $e$ into $b^{2}$ . (Note: sign is $e^{2} - 1$ , not $1 - e^{2}$ — that would give a negative $b^{2}$ .)

Step 4. Equation: $\frac{x ^{2}}{25} - \frac{y ^{2}}{144} = 1$ .

Verify: Check a focus directly: $c^{2} = a^{2} + b^{2} = 25 + 144 = 169 \Rightarrow c = 13$ ✓ (foci at $\pm 13$ ). And $2 a = 10$ matches the given difference ✓.

Example 6 — Cell F: limiting behaviour + the "is the sum constant?" check

Worked example Does the SUM stay constant? What happens far out?

For $\frac{x ^{2}}{16} - \frac{y ^{2}}{9} = 1$ ( $a = 4$ , $e = \frac{5}{4}$ ), compare $r_{1} + r_{2}$ at $x = 8$ and $x = 40$ , and describe the trend as $x \to \infty$ .

Forecast: The parent note warns the sum is $2 e x$ , not constant. Guess the sum grows as $x$ grows.

Step 1. Sum formula on the right branch: $r_{1} + r_{2} = (e x - a) + (e x + a) = 2 e x$ . Why this step? The $a$ 's cancel, isolating exactly how the sum depends on $x$ .

Step 2. At $x = 8$ : $2 e x = 2 \cdot \frac{5}{4} \cdot 8 = 20$ . At $x = 40$ : $2 e x = 2 \cdot \frac{5}{4} \cdot 40 = 100$ . Why this step? Two concrete values show the sum is genuinely changing — it is not an ellipse.

Step 3. As $x \to \infty$ , $r_{1} + r_{2} = 2 e x \to \infty$ while the difference $r_{2} - r_{1} = 2 a = 8$ stays pinned. Why this step? This is the limiting behaviour: the branch runs off to infinity, both radii blow up together, but their gap is frozen at $2 a$ . That frozen gap is the whole point of the topic.

Verify: Difference at both points: at $x = 8$ , $r_{2} - r_{1} = (10 + 4) - (10 - 4) = 8$ ; at $x = 40$ , $e x = 50$ so $(54) - (46) = 8$ . Constant $= 2 a = 8$ ✓ while the sum moved from $20$ to $100$ ✓.

Example 7 — Cell G: real-world word problem (sound timing)

Worked example Two microphones hear a bang at different times

Two microphones $S$ and $S^{'}$ sit $10$ km apart on a line. A firecracker bang reaches $S$ exactly $0.02 s$ before it reaches $S^{'}$ . Sound travels at $340 m/s$ . Show the bang lies on one branch of a hyperbola and find $a$ (in metres). Place $S = (5000, 0)$ , $S^{'} = (- 5000, 0)$ (metres).

Forecast: A fixed time difference means a fixed distance difference ( $=$ speed $\times$ time). Constant distance difference $\Rightarrow$ hyperbola. Guess the difference is a few thousand metres.

Step 1. Extra distance to the later mic: $Δ = v \cdot Δ t = 340 \times 0.02 = 6.8 m$ . Why this step? Distance $=$ speed $\times$ time; the time lead converts into how much closer the bang is to $S$ .

Step 2. $S^{'} P - S P = 6.8$ (the bang is closer to $S$ , so $S^{'}$ is farther). This constant difference equals $2 a$ . Why this step? This is literally the defining property $∣ S P - S^{'} P ∣ = 2 a$ ; the bang sits on the branch nearer $S$ .

Step 3. $2 a = 6.8 \Rightarrow a = 3.4 m$ . Why this step? Solve for $a$ from the property — the bang's locus is a hyperbola with this $a$ ; a third mic would pin the exact point (this is how sound-ranging / GPS-style location works).

Verify: Units: $340 \frac{m}{s} \times 0.02 s = 6.8 m$ (seconds cancel, metres survive) ✓. Sanity: $2 a = 6.8 <$ focal separation $10000$ , so $a < c$ as any real hyperbola needs ( $c = 5000$ , $a = 3.4$ ) ✓.

Example 8 — Cell H: exam twist (fractional difference, back out everything)

Worked example Difference given, one radius given — find the other radius and the curve

On a hyperbola with foci $(\pm 5, 0)$ , a point $P$ on the right branch has $S^{'} P = \frac{34}{3}$ (distance to the left focus). Given the constant difference of focal radii is $6$ , find $S P$ , then find $a, b, e$ and the equation.

Forecast: Difference $= 6 = 2 a$ , so $a = 3$ ; then $S P = S^{'} P - 2 a$ on the right branch. Guess $S P = \frac{34}{3} - 6 = \frac{16}{3}$ .

Step 1. $2 a = 6 \Rightarrow a = 3$ . Why this step? Given constant difference is $2 a$ by the property.

Step 2. Right branch: $S^{'} P - S P = 2 a \Rightarrow S P = S^{'} P - 2 a = \frac{34}{3} - 6 = \frac{34}{3} - \frac{18}{3} = \frac{16}{3}$ . Why this step? We use the property as a linear equation in the one unknown radius.

Step 3. Foci at $(\pm 5, 0) \Rightarrow a e = 5 \Rightarrow e = \frac{5}{3}$ . Then $b^{2} = a^{2} (e^{2} - 1) = 9 (\frac{25}{9} - 1) = 25 - 9 = 16$ . Why this step? Recover $e$ from the focus, then $b^{2}$ from the $b$ – $e$ relation to write the equation.

Step 4. Equation: $\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1$ .

Verify: Find $P$ 's $x$ : $S^{'} P = e x + a \Rightarrow \frac{34}{3} = \frac{5}{3} x + 3 \Rightarrow \frac{5}{3} x = \frac{25}{3} \Rightarrow x = 5$ . Then $S P = e x - a = \frac{5}{3} (5) - 3 = \frac{25}{3} - 3 = \frac{16}{3}$ ✓ matches Step 2. And $c^{2} = a^{2} + b^{2} = 9 + 16 = 25 \Rightarrow c = 5$ ✓ (foci at $\pm 5$ ).

Active recall

Recall Which cell does each trigger word signal?

"Distances differ by a constant" ::: build-the-curve, Cell E — that constant is $2 a$ .
A negative value pops out of $e x - a$ ::: modulus trap, Cell D/B — take absolute value.
"Time delay between two sensors" ::: real-world, Cell G — distance difference $= v Δ t = 2 a$ .
"Is the sum constant?" ::: Cell F — no, sum $= 2 e x$ , only the difference is fixed.

Recall Fast facts

Nearest-focus distance at a vertex ::: $a (e - 1)$ (e.g. Ex 3 gave $1 = 4 (\frac{5}{4} - 1)$ ). Right-branch difference vs left-branch difference ::: right: $S^{'} P - S P = 2 a$ ; left: $S P - S^{'} P = 2 a$ ; combined $∣ S P - S^{'} P ∣ = 2 a$ . How to get $b^{2}$ once $a, e$ known ::: $b^{2} = a^{2} (e^{2} - 1)$ , and independently $c^{2} = a^{2} + b^{2}$ with $c = a e$ .

Connections

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