Difference of focal radii = 2a property: Exercises

Before we start, one picture to keep the whole geometry in view.

Figure — Difference of focal radii = 2a property

The blue curve is the hyperbola. The two orange dots are the foci $S (a e, 0)$ and $S^{'} (- a e, 0)$ . For any point $P$ on it, the two green segments are the focal radii; the gap between their lengths is frozen at $2 a$ — the distance between the two red vertices.

Level 1 — Recognition

(You should be able to read numbers straight off the standard form.)

L1.1

For $\frac{x ^{2}}{16} - \frac{y ^{2}}{9} = 1$ , state $a$ , $b$ , and the constant value of $∣ S P - S^{'} P ∣$ .

Recall Solution L1.1

What we do: match the equation to $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ .

$a^{2} = 16 \Rightarrow a = 4$ . $b^{2} = 9 \Rightarrow b = 3$ .
The defining constant is $2 a === 8 ==$ .

Why: the denominator under $x^{2}$ is always $a^{2}$ for this east–west hyperbola, and the property says the difference equals $2 a$ , not $a$ .

L1.2

For $\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1$ , find the eccentricity $e$ and the coordinates of both foci.

Recall Solution L1.2

What we do: use $b^{2} = a^{2} (e^{2} - 1)$ — the hyperbola relation (note the $e^{2} - 1$ , since $e > 1$ ).

$a^{2} = 9, b^{2} = 16$ .
$16 = 9 (e^{2} - 1) \Rightarrow e^{2} - 1 = \frac{16}{9} \Rightarrow e^{2} = \frac{25}{9} \Rightarrow e = \frac{5}{3} .$
Foci at $(\pm a e, 0) = (\pm 3 \cdot \frac{5}{3}, 0) = (\pm 5, 0)$ .

Why this tool: we chose $b^{2} = a^{2} (e^{2} - 1)$ (not the ellipse's $1 - e^{2}$ ) because a hyperbola has $e > 1$ ; $e^{2} - 1$ keeps $b^{2}$ positive. See b² = a²(e²−1) relation.

Level 2 — Application

(Plug points into the focal-radius formulas.)

L2.1

For $\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1$ , compute both focal radii at the point with $x = 6$ on the right branch, and verify the property.

Recall Solution L2.1

Setup: from L1.2, $a = 3, e = \frac{5}{3}$ .

$r_{1} = ∣ e x - a ∣ = \frac{5}{3} \cdot 6 - 3 = ∣10 - 3∣ = 7.$
$r_{2} = ∣ e x + a ∣ = \frac{5}{3} \cdot 6 + 3 = ∣10 + 3∣ = 13.$
Check: $r_{2} - r_{1} = 13 - 7 = 6 = 2 a .$ ✓

Why the modulus vanishes here: on the right branch $x \geq a$ , so $e x - a \geq (e - 1) a > 0$ ; both quantities are already positive.

L2.2

For the same hyperbola, find both focal radii at $x = - 6$ (left branch). Which focus is now the farther one?

Recall Solution L2.2

What we do: same formulas, now with negative $x$ , keeping the modulus.

$e x = \frac{5}{3} (- 6) = - 10.$
$r_{1} = ∣ e x - a ∣ = ∣ - 10 - 3∣ = 13.$
$r_{2} = ∣ e x + a ∣ = ∣ - 10 + 3∣ = 7.$
Difference: $r_{1} - r_{2} = 13 - 7 = 6 = 2 a .$

Which is farther: here $r_{1} = S P = 13 > r_{2} = S^{'} P = 7$ , so the right focus $S$ is farther. On the left branch the sign of $S P - S^{'} P$ flips — that is exactly why the definition needs the absolute value.

Level 3 — Analysis

(Reason about the structure, not just plug in.)

L3.1

Show that for any point on the right branch of $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ , the sum $r_{1} + r_{2}$ is $2 e x$ and therefore not constant. What does this tell you about the difference between a hyperbola and an ellipse?

Recall Solution L3.1

What we do: add the two radii instead of subtracting.

On the right branch $r_{1} = e x - a, r_{2} = e x + a$ .
$r_{1} + r_{2} = (e x - a) + (e x + a) = 2 e x .$
This depends on $x$ : as $P$ slides out along the branch, $x$ grows, so the sum grows without bound. Not constant.

What it tells us: for the hyperbola only the difference $r_{2} - r_{1} = 2 a$ is frozen; the sum is free. For the ellipse it is the reverse — the sum is frozen. That single swap (sum ↔ difference) is the entire distinction between the two conics.

L3.2

A point on a hyperbola has focal radii $r_{1} = 4$ and $r_{2} = 10$ . The foci are $12$ apart. Find $a$ , $e$ , and $b^{2}$ .

Recall Solution L3.2

Step 1 — get $2 a$ from the property. $∣ r_{2} - r_{1} ∣ = ∣10 - 4∣ = 6 = 2 a \Rightarrow a = 3.$ Step 2 — get $a e$ from the focal separation. The foci are $S (a e, 0), S^{'} (- a e, 0)$ , distance $2 a e$ apart. So $2 a e = 12 \Rightarrow a e = 6.$ Step 3 — eccentricity. $e = \frac{a e}{a} = \frac{6}{3} = 2.$ (Indeed $e > 1$ , consistent with a hyperbola. See Eccentricity of Conics.) Step 4 — $b^{2}$ . $b^{2} = a^{2} (e^{2} - 1) = 9 (4 - 1) = 9 \cdot 3 = 27.$

Why this order: the difference of radii hands you $a$ immediately (no coordinates needed); the focal distance hands you $a e$ ; their ratio is $e$ ; then the $b$ – $e$ relation finishes it.

Level 4 — Synthesis

(Build a whole hyperbola from a description.)

L4.1

A point moves so that its distances from $(5, 0)$ and $(- 5, 0)$ always differ by $8$ . Find the equation of its path.

Recall Solution L4.1

Step 1 — identify the conic. A constant difference of distances to two fixed points is the definition of a hyperbola with those points as foci. Step 2 — read off $2 a$ . Difference $= 2 a = 8 \Rightarrow a = 4, a^{2} = 16.$ Step 3 — read off $a e$ . Foci $(\pm 5, 0) \Rightarrow a e = 5.$ Step 4 — $b^{2}$ . $b^{2} = (a e)^{2} - a^{2} = 25 - 16 = 9.$ (Why this shortcut works: $b^{2} = a^{2} (e^{2} - 1) = a^{2} e^{2} - a^{2} = (a e)^{2} - a^{2}$ .) Answer: $\frac{x ^{2}}{16} - \frac{y ^{2}}{9} = 1.$

L4.2

For $\frac{x ^{2}}{16} - \frac{y ^{2}}{9} = 1$ , find the length of the latus rectum, and the focal radii of an endpoint of the latus rectum.

Recall Solution L4.2

Step 1 — the latus rectum. It is the chord through a focus perpendicular to the major axis; its length is $\frac{2 b ^{2}}{a}$ (see Latus Rectum of a Hyperbola).

$a = 4, b^{2} = 9$ . Length $= \frac{2 \cdot 9}{4} = \frac{18}{4} = \frac{9}{2} .$ Step 2 — locate an endpoint. An endpoint of the latus rectum through $S (a e, 0)$ sits at $x = a e$ .
$e = \frac{a e}{a}$ ; here $b^{2} = a^{2} (e^{2} - 1) \Rightarrow 9 = 16 (e^{2} - 1) \Rightarrow e^{2} = \frac{25}{16} \Rightarrow e = \frac{5}{4}$ , so $a e = 4 \cdot \frac{5}{4} = 5$ , and $x = 5$ . Step 3 — focal radii at $x = 5$ (right branch).
$r_{1} = ∣ e x - a ∣ = \frac{5}{4} \cdot 5 - 4 = \frac{25}{4} - 4 = \frac{9}{4} .$
$r_{2} = ∣ e x + a ∣ = \frac{25}{4} + 4 = \frac{41}{4} .$
Check property: $r_{2} - r_{1} = \frac{41}{4} - \frac{9}{4} = \frac{32}{4} = 8 = 2 a .$ ✓
Sanity on $r_{1}$ : the endpoint is directly above the focus, so its shorter focal radius equals half the latus rectum $= \frac{9}{4}$ . It matches. ✓

Level 5 — Mastery

(Everything at once, plus a limiting case.)

L5.1

A hyperbola has one vertex at $(3, 0)$ and passes through $(6, 23)$ , centred at the origin with foci on the $x$ -axis. Find its equation and confirm $2 a$ equals the difference of focal radii at $(6, 23)$ .

Recall Solution L5.1

Step 1 — $a$ from the vertex. Vertex at $(a, 0) = (3, 0) \Rightarrow a = 3, a^{2} = 9.$ Step 2 — $b^{2}$ from the point. Plug $(6, 23)$ into $\frac{x ^{2}}{9} - \frac{y ^{2}}{b ^{2}} = 1$ : $\frac{36}{9} - \frac{12}{b ^{2}} = 1 \Rightarrow 4 - \frac{12}{b ^{2}} = 1 \Rightarrow \frac{12}{b ^{2}} = 3 \Rightarrow b^{2} = 4.$ Equation: $\frac{x ^{2}}{9} - \frac{y ^{2}}{4} = 1.$ Step 3 — get $e$ . $b^{2} = a^{2} (e^{2} - 1) \Rightarrow 4 = 9 (e^{2} - 1) \Rightarrow e^{2} = \frac{13}{9} \Rightarrow e = \frac{13}{3} .$ Step 4 — radii at $(6, 23)$ (right branch, $x = 6$ ). Note $e x = \frac{13}{3} \cdot 6 = 213$ .

$r_{1} = ∣ e x - a ∣ = 213 - 3.$
$r_{2} = ∣ e x + a ∣ = 213 + 3.$
Difference: $r_{2} - r_{1} = (213 + 3) - (213 - 3) = 6 = 2 a .$ ✓
(Cross-check by raw distance: foci at $(\pm a e, 0) = (\pm 213, 0)$ ; $S P = (6 - 213)^{2} + 12$ , $S^{'} P = (6 + 213)^{2} + 12$ , and $S^{'} P - S P = 6$ .)

L5.2

As a point $P$ on the right branch slides toward the vertex $(a, 0)$ , what happens to each focal radius, and does the difference $2 a$ survive at the vertex itself?

Recall Solution L5.2 (limiting / degenerate case)

What we do: take the limit $x \to a^{+}$ in $r_{1} = e x - a, r_{2} = e x + a$ .

At the vertex $x = a$ : $r_{1} = e a - a = a (e - 1)$ and $r_{2} = e a + a = a (e + 1)$ .
Both are the shortest possible on this branch (the vertex is the closest point to $S$ ).
Difference: $r_{2} - r_{1} = a (e + 1) - a (e - 1) = 2 a .$ Still exactly $2 a$ .

Geometric reading: at the vertex the point sits on the axis between the foci. Then $S P + S^{'} P$ equals the full focus-to-focus span minus/plus nothing along the line: $S^{'} P = a e + a$ , $S P = a e - a$ , and their gap is $2 a$ — the vertex-to-vertex distance. The property holds even at this extreme point; it never breaks down.

Active recall

Recall One-line self-test

From $\frac{x ^{2}}{16} - \frac{y ^{2}}{9} = 1$ , what is the constant difference of focal radii? ::: $2 a = 8$ .
Radii at $x = 6$ on $\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1$ ? ::: $r_{1} = 7, r_{2} = 13$ .
Radii $4$ and $10$ , foci $12$ apart — find $e$ . ::: $a = 3, a e = 6, e = 2$ .
Locus: distances to $(\pm 5, 0)$ differ by $8$ ? ::: $\frac{x ^{2}}{16} - \frac{y ^{2}}{9} = 1$ .
At the vertex, difference of radii? ::: $a (e + 1) - a (e - 1) = 2 a$ , unchanged.

Connections

3.4.08 Difference of focal radii = 2a property (Hinglish)
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