Hyperbola — standard forms, asymptotes, foci, eccentricity

3.4.7 · Maths › Conic Sections

Intuition Badi tasveer

Ek ellipse un points ka set hota hai jinki do foci se distances add hokar ek constant banati hain. Hyperbola is operation ko palat deta hai: yeh un points ka set hai jinki do foci se distances ka fark ek constant hota hai. Bas yeh ek sign change ( $+ \to -$ ) hi wajah hai ki hyperbola do open branches mein split ho jaata hai jo infinity ki taraf bhaagti hain, band nahi hoti.

WHY it exists: ek double cone ko steeply kaatna (side se bhi teekhaa) dono nappes ko slice karta hai → do branches.
WHAT it gives us: open curves jiske saath straight-line guides (asymptotes) hote hain jinhe curve follow karta hai par kabhi touch nahi karta.
HOW we use it: unbound comets ki orbits, LORAN navigation (constant time-difference = constant distance-difference), aur lens/mirror design.

1. Standard equation ko scratch se derive karna

Setup (WHY this coordinate choice): foci ko $x$ -axis par symmetrically $F_{1} (- c, 0)$ aur $F_{2} (c, 0)$ par rakho taaki algebra symmetric rahe aur centre origin par ho.

Maano $P = (x, y)$ . Tab $(x + c)^{2} + y^{2} - (x - c)^{2} + y^{2} = \pm 2 a .$

Step 1 — ek root ko isolate karo. Kyun? Ek baar mein ek square root manage karna aasaan hota hai. $(x + c)^{2} + y^{2} = \pm 2 a + (x - c)^{2} + y^{2} .$

Step 2 — dono sides ko square karo. Kyun? Isolated radical khatam ho jaata hai. $(x + c)^{2} + y^{2} = 4 a^{2} \pm 4 a (x - c)^{2} + y^{2} + (x - c)^{2} + y^{2} .$

Expand karo aur $x^{2}, c^{2}, y^{2}$ cancel karo: $4 c x = 4 a^{2} \pm 4 a (x - c)^{2} + y^{2} \Rightarrow c x - a^{2} = \pm a (x - c)^{2} + y^{2} .$

Step 3 — phir se square karo. Kyun? Baaki ka radical bhi khatam ho jaata hai. $c^{2} x^{2} - 2 a^{2} c x + a^{4} = a^{2} [(x - c)^{2} + y^{2}] = a^{2} x^{2} - 2 a^{2} c x + a^{2} c^{2} + a^{2} y^{2} .$

$- 2 a^{2} c x$ cancel karo, regroup karo: $(c^{2} - a^{2}) x^{2} - a^{2} y^{2} = a^{2} (c^{2} - a^{2}) .$

Step 4 — bacha hua naam do. Kyunki $c > a$ hai, quantity $c^{2} - a^{2} > 0$ hai. Define karo: $b^{2} = c^{2} - a^{2} .$ $b$ define kyun karo? Yeh equation ko clean banata hai AUR baad mein asymptote slopes fix karne waala vertical semi-extent nikalta hai. $a^{2} b^{2}$ se divide karo:

Vertical version bas roles swap karta hai: $\frac{y ^{2}}{a ^{2}} - \frac{x ^{2}}{b ^{2}} = 1 (foci on the y -axis, (0, \pm c)) .$

2. Eccentricity — shape number

$c^{2} = a^{2} + b^{2}$ use karke: $e = \frac{a ^{2} + b ^{2}}{a} = 1 + \frac{b ^{2}}{a ^{2}} .$

Directrix form: har focus ek directrix line $x = \pm a / e$ se pair hota hai, aur har point $P$ ke liye: $\frac{P F}{(distance to that directrix)} = e .$ Yeh sabhi conics ki single unifying definition hai.

3. Asymptotes — straight-line guides derive karna

Derivation. $y$ ke liye solve karo: $y^{2} = b^{2} (\frac{x ^{2}}{a ^{2}} - 1) \Rightarrow y = \pm \frac{b}{a} x 1 - \frac{a ^{2}}{x ^{2}} .$ Jab $∣ x ∣ \to \infty$ , $1 - a^{2} / x^{2} \to 1$ , isliye $y \to \pm \frac{b}{a} x .$

Gap → 0 proof (Kyun kabhi cross nahi karte): curve aur line ke beech vertical gap hai $\frac{b}{a} x - \frac{b}{a} x 1 - \frac{a ^{2}}{x ^{2}} = \frac{b}{a} x (1 - 1 - \frac{a ^{2}}{x ^{2}}) x \to \infty 0,$ kyunki bracket $\frac{a ^{2}}{2 x ^{2}}$ ki tarah shrink karta hai aur $x$ sirf linearly badhta hai. Product $\to 0$ : curve line ko hug karta hai par kabhi nahi milta.

Figure — Hyperbola — standard forms, asymptotes, foci, eccentricity

4. Rectangular (equilateral) hyperbola

Worked Examples

Flashcards

Standard hyperbola equation (transverse on x)

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

Hyperbola ke liye focus relation

c^{2} = a^{2} + b^{2}

(foci vertices ke bahar hote hain)

Hyperbola ki eccentricity aur uska range

e = c / a

, aur

e > 1

hamesha

a aur b ke terms mein eccentricity

e = 1 + b^{2} / a^{2}

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

ke asymptotes

y = \pm (b / a) x

Hyperbola ki defining locus property

∣ P F_{1} - P F_{2} ∣ = 2 a

, focal distances ka constant fark

Transverse vs conjugate axis ki length

transverse

= 2 a

(foci iske andar), conjugate

= 2 b

Rectangular hyperbola kya hota hai?

a = b

; asymptotes

y = \pm x

perpendicular;

e = 2

Hyperbola ki directrix

lines

x = \pm a / e

, jahan

P F / (dist to directrix) = e

Ellipse vs hyperbola focus relation

ellipse

c^{2} = a^{2} - b^{2}

(minus); hyperbola

c^{2} = a^{2} + b^{2}

(plus)

Asymptotes kyun appear hote hain?

door jaane par "

1

" negligible ho jaata hai isliye curve →

x^{2} / a^{2} - y^{2} / b^{2} = 0

Recall Feynman: ek 12-saal ke bachche ko samjhao

Socho do thumbtacks (foci) hain aur tum is tarah chalo ki ek tack hamesha exactly, maano, 4 kadam kareeb ho doosre se. Total nahi — fark. Agar yeh fark fixed rakho aur trace karo ki tum kahan khadhe ho sakte ho, tumhe ek aisi curve milti hai jo ek tack ki taraf jhaanke, phir ek mirror copy doosri tack ki taraf jhaanke: do alag arcs. Door edges par yeh arcs seedhi ho jaati hain aur almost perfectly do crossing straight rulers (asymptotes) ke saath-saath chalti hain — hamesha ke liye kareeb aati rehti hain par kabhi touch nahi karti, jaise ek car ek diwar ke saath daud rahi ho jo kabhi scrape nahi hoti. Yeh arcs kitni "wide open" hain yeh ek number se batata hai, $e$ ; hyperbola ke liye $e$ hamesha 1 se bada hota hai.

Connections

Ellipse — standard forms, foci, eccentricity — same derivation jahan $+ \to -$ aur $c^{2} = a^{2} - b^{2}$ .
Parabola — standard forms, focus, directrix — borderline case $e = 1$ .
Eccentricity — unified conic definition — ek focus–directrix ratio sabhi conics classify karta hai.
Asymptotes and Limits at Infinity — gap-goes-to-zero argument ek limit hai.
Rectangular Hyperbola xy=c^2 — $a = b$ case ka $4 5^{\circ}$ rotation.
Conic Sections from a Double Cone — dono nappes se guzarta steep cut.