Difference of focal radii = 2a property

WHAT is a focal radius?

WHY two foci? A hyperbola is genuinely defined by both foci — you cannot describe it with one. The defining relation is: $|SP - S'P| = \text{constant} = 2a$

HOW: Derive that the difference is $2a$ (from scratch)

We prove the standard equation forces the difference to be $2a$ . Start from the curve, land on the property.

Step 1 — Set up the focal radii. Take $P(x,y)$ on the right branch. Foci $S(ae,0)$ , $S'(-ae,0)$ . $r_1 = SP = \sqrt{(x-ae)^2+y^2}, \qquad r_2 = S'P=\sqrt{(x+ae)^2+y^2}$ Why this step? Distances are just Pythagoras from each focus; nothing assumed yet.

Step 2 — Kill the square root using the curve equation. From $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ we get $y^2 = b^2\!\left(\dfrac{x^2}{a^2}-1\right)$ . Substitute with $b^2=a^2(e^2-1)$ : $r_2^2 = (x+ae)^2 + a^2(e^2-1)\left(\tfrac{x^2}{a^2}-1\right)$ $= x^2+2aex+a^2e^2 + (e^2-1)x^2 - a^2(e^2-1)$ Why this step? Replacing $y^2$ removes $y$ , leaving a pure function of $x$ that we can hope to make a perfect square.

Step 3 — Collect and factor. $r_2^2 = e^2x^2 + 2aex + a^2 = (ex+a)^2$ So $r_2 = |ex+a|$ . On the right branch $x\ge a>0$ and $e>1$ , so $ex+a>0$ : $\boxed{r_2 = S'P = ex+a}$ Why this step? The mess collapsed to a perfect square — that's the payoff of using $b^2=a^2(e^2-1)$ .

Step 4 — Do the same for $r_1$ . By identical algebra with the $-ae$ focus: $r_1^2 = (ex-a)^2 \Rightarrow r_1 = |ex-a| = ex-a \quad(\text{since } ex\ge ea>a)$ $\boxed{r_1 = SP = ex-a}$

Step 5 — Take the difference. $r_2 - r_1 = (ex+a)-(ex-a) = 2a$ Why this step? The $ex$ terms cancel — the $x$ -dependence vanishes, so the difference is the same for every point on the branch. That constancy is the whole property. $\blacksquare$

Figure — Difference of focal radii = 2a property

WHY the absolute value / two branches

On the right branch $S'$ (left focus) is farther, so $S'P > SP$ and $S'P-SP=2a$ . On the left branch $S$ (right focus) is farther, so $SP-S'P=2a$ . The sign flips between branches, hence $|SP-S'P|=2a$ captures both. This is why a hyperbola has two disconnected branches, unlike an ellipse.

Worked examples

Worked example Example 1 — Compute both radii

For $\dfrac{x^2}{9}-\dfrac{y^2}{16}=1$ , find the focal radii at $x=5$ (right branch). Step 1: $a^2=9\Rightarrow a=3$ , $b^2=16$ . Why? Read straight off standard form. Step 2: $b^2=a^2(e^2-1)\Rightarrow 16=9(e^2-1)\Rightarrow e^2=\tfrac{25}{9}\Rightarrow e=\tfrac53$ . Why? This links $b$ to $e$ . Step 3: $r_1 = ex-a = \tfrac53(5)-3 = \tfrac{25}{3}-3=\tfrac{16}{3}$ . $r_2 = ex+a=\tfrac{25}{3}+3=\tfrac{34}{3}$ . Step 4 (check): $r_2-r_1=\tfrac{34}{3}-\tfrac{16}{3}=\tfrac{18}{3}=6=2a.$ ✓ Why check? Confirms the property numerically.

Common mistakes

Active recall

Recall Try before revealing

What quantity is constant on a hyperbola, and equal to what?
Write the two focal radii of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ .
Why is there an absolute value in $|SP-S'P|=2a$ ?
Is $r_1+r_2$ constant for a hyperbola?

Answers: difference of focal radii $=2a$ ; $r_1=|ex-a|,\,r_2=|ex+a|$ ; because the far focus swaps between branches flipping the sign; No, $r_1+r_2=2ex$ varies.

Recall Feynman: explain to a 12-year-old

Imagine two friends, Sam and Sue, standing on a field. You walk around so that no matter where you stand, Sue is always exactly 6 metres more distant to you than Sam — but you swap "which is closer" on the two sides. The path you trace is a hyperbola, and that steady gap "6 metres" is the width between the curve's two tips ( $2a$ ). An ellipse is the opposite game: keep the total walk to both friends the same, and you get an oval.

Flashcards

Hyperbola defining property (distances)

The absolute difference of distances from any point to the two foci is constant,

|SP-S'P|=2a

Focal radii of

\frac{x^2}{a^2}-\frac{y^2}{b^2}=1

r_1=SP=|ex-a|

r_2=S'P=|ex+a|

Why does the difference come out constant?

r_2-r_1=(ex+a)-(ex-a)=2a

; the

ex

(x-dependence) cancels.

b

–

e

relation for a hyperbola

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

with

e>1

Is the SUM of focal radii constant for a hyperbola?

No;

r_1+r_2=2ex

, which depends on

x

Why the modulus in the definition?

On different branches the farther focus swaps, flipping the sign of

SP-S'P

Locus of points with distances to

(\pm4,0)

differing by 6

Hyperbola

\frac{x^2}{9}-\frac{y^2}{7}=1

(

a=3,ae=4,b^2=7

Ellipse vs hyperbola in one word

Ellipse fixes the SUM, hyperbola fixes the DIFFERENCE.

Connections

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

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, hyperbola ka asli definition bahut simple hai: koi bhi point lo curve pe, aur uski do foci (S aur S') se distance nikaalo. In dono distances ka difference hamesha constant rehta hai, aur wo constant exactly $2a$ hota hai — yaani do vertices ke beech ki doori. Ellipse mein sum fix hota hai, hyperbola mein difference fix hota hai. Bas yahi ek shabd ka farq poora khel badal deta hai.

Derivation ka core idea: focal radii $r_1=\sqrt{(x-ae)^2+y^2}$ aur $r_2=\sqrt{(x+ae)^2+y^2}$ . Curve ki equation se $y^2$ ki value daalo aur $b^2=a^2(e^2-1)$ use karo — toh sab kuch perfect square ban jaata hai: $r_1=ex-a$ , $r_2=ex+a$ . Ab difference lo: $r_2-r_1=2a$ . Dekha? $ex$ waala part cancel ho gaya, isliye har point pe answer same $2a$ aata hai. Yehi constancy hi property hai.

Ek important cheez yaad rakho — sum ( $r_1+r_2=2ex$ ) constant nahi hota, wo $x$ ke saath change hota hai. Isliye agar exam mein "sum" waala confuse kara de toh phas jaoge. Rule yaad rakho: equation mein minus sign hai, toh definition mein bhi difference aayega. Aur $b^2=a^2(e^2-1)$ use karna, ellipse waala $a^2(1-e^2)$ nahi — kyunki hyperbola mein $e>1$ hota hai.

Practical use: agar koi bole "point ki do fixed points se distance ka farq constant hai", turant samajh jao ki locus hyperbola hai, aur $2a$ = wo constant. Foci se $ae$ nikaalo, phir $b^2$ , aur equation ready. Yeh trick numerical questions mein seconds mein answer de deti hai.

Difference of focal radii = 2a property

WHAT is a focal radius?

HOW: Derive that the difference is $2a$ (from scratch)

WHY the absolute value / two branches

Worked examples

Common mistakes

Active recall

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Conic Sections

Connections