Difference of focal radii = 2a property: Question bank

Before you start, one anchor so every symbol below is earned:

$S (a e, 0)$ and $S^{'} (- a e, 0)$ are the two foci (fixed points). Here $a$ is half the distance between the two vertices — the two sharp turning points where each branch is nearest the centre (informally, its "tips"), sitting at $(\pm a, 0)$ . The number $e$ (the eccentricity, always $> 1$ for a hyperbola) measures how "open" the curve is — see Eccentricity of Conics.
$b$ is the second scale of the standard hyperbola $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ ; it controls the slope of the asymptotes and is tied to $a$ and $e$ by $b^{2} = a^{2} (e^{2} - 1)$ — see b² = a²(e²−1) relation.
For a point $P$ on the curve, $S P$ and $S^{'} P$ are its focal radii — the two straight distances from $P$ to each focus.
The property: $∣ S P - S^{'} P ∣ = 2 a$ . The working formulas $r_{1} = S P = ∣ e x - a ∣$ and $r_{2} = S^{'} P = ∣ e x + a ∣$ come from substituting $y^{2} = b^{2} (x^{2} / a^{2} - 1)$ with $b^{2} = a^{2} (e^{2} - 1)$ into $S P^{2}$ and $S^{'} P^{2}$ , which collapses each into a perfect square — full derivation in the parent note here. On the right branch ( $x \geq a$ ) the moduli drop to $r_{1} = e x - a$ , $r_{2} = e x + a$ ; on the left branch ( $x \leq - a$ ) the signs inside the moduli flip.

The picture below is the mental image to hold for every question on this page — the two branches, the two foci farther out than the vertices, and the same gap $2 a$ appearing on both sides.

Figure — Difference of focal radii = 2a property

And this one shows why the difference stays pinned while the individual radii blow up as the point runs to infinity — keep it in mind for the "edge cases" section.

True or false — justify

Is the SUM

S P + S^{'} P

constant on a hyperbola?

False. On the right branch (

x \geq a

)

S P + S^{'} P = (e x - a) + (e x + a) = 2 e x

; on the left branch (

x \leq - a

) both moduli flip so

S P + S^{'} P = - 2 e x

. Either way it depends on

x

— only the difference cancels the

e x

term. The constant sum belongs to the ellipse.

"Both branches together give

∣ S P - S^{'} P ∣ = 2 a

." True or false?

True. On the right branch

S^{'} P - S P = 2 a

; on the left branch

S P - S^{'} P = 2 a

. The far focus swaps sides, so the absolute value covers both cases at once.

Using the two-foci definition (constant difference of distances to two fixed points), a hyperbola can be drawn with a single focus. True or false?

False for the two-foci definition: you need two fixed points to subtract distances, so one focus gives nothing to difference. (Note that a different definition — the focus–directrix one — does use a single focus plus a directrix line; that is a separate construction, not the one this property is about.)

For every point on a hyperbola,

S P - S^{'} P

equals

+ 2 a

. True or false?

False. It equals

+ 2 a

on one branch and

- 2 a

on the other; only

∣ S P - S^{'} P ∣ = 2 a

holds everywhere.

The constant

2 a

equals the distance between the two vertices. True or false?

True. The vertices sit at

(\pm a, 0)

, so they are

2 a

apart, which is exactly the fixed difference.

Increasing

e

(making the hyperbola more open) changes the constant difference

2 a

. True or false?

False. The difference is fixed at

2 a

regardless of

e

;

e

only moves the foci (

a e

) and reshapes the branches, not the vertex gap.

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

holds for a hyperbola." True or false?

False — that is the ellipse relation (

e < 1

). For a hyperbola

e > 1

, so we must use

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

to keep

b^{2} > 0

; see b² = a²(e²−1) relation.

A focal radius of a hyperbola can be negative. True or false?

False. A focal radius is a genuine distance, so it is always

\geq 0

; the formulas carry a modulus,

r_{1} = ∣ e x - a ∣

, precisely to protect that.

Spot the error

r_{1} = e x - a

, so at

x = 0

we get

r_{1} = - a

." What went wrong?

The unsigned form

r_{1} = e x - a

only holds on the right branch (

x \geq a

); the general formula is

r_{1} = ∣ e x - a ∣

, and no real hyperbola point has

x = 0

anyway (the branches start at

∣ x ∣ \geq a

). Plugging

x = 0

describes a point off the curve, so the formula does not apply there.

"Since ellipse and hyperbola have twin equations, both fix the SUM to

2 a

." Find the flaw.

The equations differ by a sign: ellipse has

+ \frac{y ^{2}}{b ^{2}}

, hyperbola has

- \frac{y ^{2}}{b ^{2}}

. That minus in the equation mirrors the difference (not sum) in the definition — sum on a hyperbola is

\pm 2 e x

, not constant.

A student writes

∣ S P - S^{'} P ∣ = 2 a

but then says "so

S P

and

S^{'} P

are each

a

." Spot the error.

A constant difference says nothing about the individual radii;

S P

and

S^{'} P

each vary with position (they are

∣ e x \mp a ∣

). Only their gap stays fixed at

2 a

S P - S^{'} P = 2 a

on the right branch." Correct the sign.

On the right branch the left focus

S^{'}

is farther, so

S^{'} P > S P

, giving

S^{'} P - S P = + 2 a

. Writing

S P - S^{'} P

there yields

- 2 a

, so the labels are reversed.

"The foci are at

(\pm a, 0)

." What's wrong?

Those are the vertices. The foci sit farther out at

(\pm a e, 0)

with

e > 1

, so

a e > a

— the foci always lie beyond the vertices.

"For

\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1

, the difference of radii is

2 b = 8

." Fix it.

The constant difference is

2 a

, not

2 b

. Here

a^{2} = 9 \Rightarrow a = 3

, so the difference is

2 a = 6

;

b

never enters the focal-difference value.

Why questions

Why does the

e x

term cancel when we subtract the two focal radii?

On one branch both radii share the same

e x

piece (

e x + a

and

e x - a

), so subtraction removes it, leaving only

2 a

. That cancellation is why the difference is position-independent.

Why is an absolute value needed in

∣ S P - S^{'} P ∣ = 2 a

Because which focus is farther flips between the two branches, flipping the sign of

S P - S^{'} P

; the modulus collapses

+ 2 a

and

- 2 a

into one clean statement.

Why does a hyperbola have two disconnected branches while an ellipse is one closed loop?

A difference of distances splits points into "closer to

S

" and "closer to

S^{'}

" groups, forming two arms; a sum traps every point inside one bounded oval, so the ellipse stays connected.

Why must

e > 1

for the difference construction to give a real curve?

We need the foci (

a e

) farther apart than the vertices (

a

), i.e.

a e > a

, so

e > 1

. If

e = 1

the foci meet the vertices and the curve degenerates;

e < 1

would place foci inside and give an ellipse instead.

Why does the substitution

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

make the focal-radius algebra collapse to a perfect square?

Start with

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

and put

y^{2} = b^{2} (x^{2} / a^{2} - 1)

. With

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

the

y^{2}

term becomes

(e^{2} - 1) x^{2} - a^{2} (e^{2} - 1)

; adding

(x + a e)^{2} = x^{2} + 2 a e x + a^{2} e^{2}

gives

e^{2} x^{2} + 2 a e x + a^{2} = (e x + a)^{2}

. The chosen relation is exactly what makes those pieces line up as a square.

Why is

r_{1} + r_{2} = \pm 2 e x

actually useful even though it is not constant?

Its magnitude grows linearly with

∣ x ∣

(it is

2 e ∣ x ∣

), so a distances-sum that scales with position flags a hyperbola, whereas a constant sum flags an ellipse — a one-glance test, provided you take the absolute value since the raw sign flips between branches.

Edge cases

At a vertex

(a, 0)

, what are the two focal radii and their difference?

r_{1} = ∣ e \cdot a - a ∣ = a (e - 1)

and

r_{2} = ∣ e \cdot a + a ∣ = a (e + 1)

; their difference is

a (e + 1) - a (e - 1) = 2 a

. The property holds even at the vertex.

What happens to the two focal radii as

P

races off to infinity along a branch?

Both

S P

and

S^{'} P

grow without bound (roughly like

e ∣ x ∣

), yet their difference stays pinned at exactly

2 a

— the gap never widens no matter how far out you go.

Can the constant difference ever be

0

, and what curve would that be?

2 a = 0

the "hyperbola" degenerates to the perpendicular bisector line of the two foci (all points equidistant), not a genuine hyperbola — so we always need

a > 0

Is the point where

S P = S^{'} P

ever on the hyperbola?

No.

S P = S^{'} P

forces

S P - S^{'} P = 0 \neq = 2 a

, so such points lie on the axis of symmetry (the

y

-axis), which the branches never touch.

If the given difference

2 a

exceeds the distance

2 a e

between the foci, is a hyperbola possible?

No. We need

2 a < 2 a e

, i.e.

e > 1

; if the required difference matched or beat the focal separation, no point could satisfy it (the triangle inequality forbids

∣ S P - S^{'} P ∣ > S S^{'}

Connections

Difference of focal radii = 2a property (index 3.4.8)
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