Difference of focal radii = 2a property — Hinglish notes

3.4.8 · Maths › Conic Sections

WHAT hai ek focal radius?

WHY do foci? Hyperbola genuinely dono foci se define hota hai — isse ek focus se describe nahi kar sakte. Defining relation yeh hai: $∣ S P - S^{'} P ∣ = constant = 2 a$

HOW: Prove karo ki difference $2 a$ hai (scratch se)

Hum prove karte hain ki standard equation difference ko $2 a$ hone par force karti hai. Curve se shuru karo, property par land karo.

Step 1 — Focal radii set up karo. $P (x, y)$ lo right branch par. Foci $S (a e, 0)$ , $S^{'} (- a e, 0)$ . $r_{1} = S P = (x - a e)^{2} + y^{2}, r_{2} = S^{'} P = (x + a e)^{2} + y^{2}$ Yeh step kyun? Distances sirf Pythagoras hain har focus se; abhi kuch assume nahi kiya.

Step 2 — Square root ko curve equation use karke khatam karo. $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ se milta hai $y^{2} = b^{2} (\frac{x ^{2}}{a ^{2}} - 1)$ . $b^{2} = a^{2} (e^{2} - 1)$ ke saath substitute karo: $r_{2}^{2} = (x + a e)^{2} + a^{2} (e^{2} - 1) (\frac{x ^{2}}{a ^{2}} - 1)$ $= x^{2} + 2 a e x + a^{2} e^{2} + (e^{2} - 1) x^{2} - a^{2} (e^{2} - 1)$ Yeh step kyun? $y^{2}$ replace karne se $y$ hat jaata hai, aur ek pure function of $x$ bachta hai jise hum perfect square banana ki hope karte hain.

Step 3 — Collect karo aur factor karo. $r_{2}^{2} = e^{2} x^{2} + 2 a e x + a^{2} = (e x + a)^{2}$ Toh $r_{2} = ∣ e x + a ∣$ . Right branch par $x \geq a > 0$ aur $e > 1$ , isliye $e x + a > 0$ : $r_{2} = S^{'} P = e x + a$ Yeh step kyun? Saari gandagi ek perfect square mein collapse ho gayi — yahi payoff hai $b^{2} = a^{2} (e^{2} - 1)$ use karne ka.

Step 4 — $r_{1}$ ke liye bhi same karo. $- a e$ focus ke saath identical algebra se: $r_{1}^{2} = (e x - a)^{2} \Rightarrow r_{1} = ∣ e x - a ∣ = e x - a (since e x \geq e a > a)$ $r_{1} = S P = e x - a$

Step 5 — Difference lo. $r_{2} - r_{1} = (e x + a) - (e x - a) = 2 a$ Yeh step kyun? $e x$ terms cancel ho jaate hain — $x$ -dependence vanish ho jaati hai, toh difference branch par har point ke liye same hai. Yahi constancy poori property hai. $■$

Figure — Difference of focal radii = 2a property

WHY absolute value / do branches

Right branch par $S^{'}$ (left focus) door hai, toh $S^{'} P > S P$ aur $S^{'} P - S P = 2 a$ . Left branch par $S$ (right focus) door hai, toh $S P - S^{'} P = 2 a$ . Sign branches ke beech flip hota hai, isliye $∣ S P - S^{'} P ∣ = 2 a$ dono ko capture karta hai. Isi liye hyperbola ke do disconnected branches hote hain, ellipse ke unlike.

Worked examples

Common mistakes

Active recall

Recall Reveal karne se pehle try karo

Hyperbola par kaun si quantity constant hoti hai, aur kis cheez ke barabar?
$\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ ke do focal radii likho.
$∣ S P - S^{'} P ∣ = 2 a$ mein absolute value kyun hai?
Kya $r_{1} + r_{2}$ hyperbola ke liye constant hai?

Answers: focal radii ka difference $= 2 a$ ; $r_{1} = ∣ e x - a ∣, r_{2} = ∣ e x + a ∣$ ; kyunki door wala focus branches ke beech swap hota hai sign flip karke; Nahi, $r_{1} + r_{2} = 2 e x$ vary karta hai.

Recall Feynman: ek 12-saal ke bachche ko explain karo

Socho do dost hain, Sam aur Sue, ek field mein khade hain. Tum idhar-udhar chalte ho taaki chahe kahan bhi khado, Sue hamesha exactly 6 metre zyada door ho tumse compared to Sam — lekin "kaun paas hai" yeh dono sides par swap hota hai. Jo path tum trace karte ho woh hyperbola hai, aur woh steady gap "6 metre" curve ke dono tips ke beech ki width hai ( $2 a$ ). Ellipse ulta game hai: dono doston tak ki total walk same rakho, aur tumhe ek oval milta hai.

Flashcards

Hyperbola defining property (distances)

Kisi bhi point ki do foci se distances ka absolute difference constant hota hai,

∣ S P - S^{'} P ∣ = 2 a

Focal radii of

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

r_{1} = S P = ∣ e x - a ∣

r_{2} = S^{'} P = ∣ e x + a ∣

Difference constant kyun nikalta hai?

r_{2} - r_{1} = (e x + a) - (e x - a) = 2 a

;

e x

(x-dependence) cancel ho jaata hai.

b

–

e

relation for a hyperbola

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

with

e > 1

Kya hyperbola ke liye focal radii ka SUM constant hai?

Nahi;

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

, jo

x

par depend karta hai.

Definition mein modulus kyun?

Alag branches par door wala focus swap hota hai,

S P - S^{'} P

ka sign flip karta hai.

Points ka locus jab

(\pm 4, 0)

se distances

6

differ karti hain

Hyperbola

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

(

a = 3, a e = 4, b^{2} = 7

Ellipse vs hyperbola ek word mein

Ellipse SUM fix karta hai, hyperbola DIFFERENCE fix karta hai.

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