Rectangular hyperbola xy = c² — Hinglish notes

3.4.9 · Maths › Conic Sections

WHY does $x^{2} - y^{2} = a^{2}$ become $x y = c^{2}$ ?

Derivation (first principles se — axes rotate karo). Shuru karo $x^{2} - y^{2} = a^{2}$ se. Coordinate axes ko $4 5^{\circ}$ rotate karo. Naye frame mein ek point $(X, Y)$ ka purane $(x, y)$ se relation: $x = X cos 4 5^{\circ} - Y sin 4 5^{\circ} = \frac{X - Y}{2}, y = X sin 4 5^{\circ} + Y cos 4 5^{\circ} = \frac{X + Y}{2}$

Yeh step kyun? Rotation distances/shape ko preserve karta hai; hum sirf unhi points ko turned axes mein relabel kar rahe hain.

Substitute karo: $x^{2} - y^{2} = \frac{( X - Y ) ^{2}}{2} - \frac{( X + Y ) ^{2}}{2} = \frac{- 4 X Y}{2} = - 2 X Y$ Toh $- 2 X Y = a^{2}$ , yaani $X Y = - \frac{a ^{2}}{2}$ . $X Y = x^{'} y^{'}$ rename karke aur sign / labels is tarah choose karke ki branches quadrants I aur III mein rahe: $x y = c^{2}, c^{2} = \frac{a ^{2}}{2}$

Yeh step kyun? Yeh dikhata hai ki $x y = c^{2}$ koi nayi curve nahi — yeh $x^{2} - y^{2} = a^{2}$ hi hai, bas axes asymptotes ke saath choose ki gayi hain ( $x = 0$ aur $y = 0$ ab asymptotes hain).

Parametric point — asli kaam ka tool

Ek point par slope (tangent derive karo). $x y = c^{2}$ ko implicitly differentiate karo: $y + x \frac{d y}{d x} = 0 \Rightarrow \frac{d y}{d x} = - \frac{y}{x}$ .

$(c t, c / t)$ par: slope $= - \frac{c / t}{c t} = - \frac{1}{t ^{2}}$ .

Tangent equation (point-slope): $y - \frac{c}{t} = - \frac{1}{t ^{2}} (x - c t)$ $t^{2}$ se multiply karo: $t^{2} y - c t = - x + c t \Rightarrow x + t^{2} y = 2 c t$ , yaani $\frac{x}{t} + t y = 2 c$

Yeh step kyun? $t$ se divide karne par ek yaadgaar symmetric form milti hai.

Normal equation (slope $= + t^{2}$ , negative reciprocal): $y - \frac{c}{t} = t^{2} (x - c t) \Rightarrow x t^{3} - y t - c t^{4} + c = 0$

Worked Examples

Recall Feynman: 12 saal ke bachche ko samjhao

Socho do bilkul seedhi sadakein hain jo right angles par cross karti hain (x aur y axes). Ab ek curve banao jo dono sadakon ke karib rehe par unhe kabhi na chhue, do opposite corners mein. Curve par jahan bhi khado, dono sadakon ki taraf jo box banta hai uska area hamesha same rehta hai — woh fixed area $c^{2}$ hai. Yahi "same-area rule" hi curve $x y = c^{2}$ hai. Yeh asliyat mein ordinary hyperbola hi hai, bas 45° sar jhukake dekhi gayi.

Flashcards

Hyperbola "rectangular" kyun kehlaata hai?

Iske asymptotes perpendicular hote hain (equivalently

a = b

e = 2

Rectangular hyperbola ki eccentricity?

e = 2

Rectangular hyperbola ka standard form jab asymptotes ke saath refer kiya jaaye?

x y = c^{2}

x y = c^{2}

par parametric point?

(c t, c / t)

x y = c^{2}

ke liye

d y / d x

- y / x

;

(c t, c / t)

par yeh

- 1/ t^{2}

hai.

x y = c^{2}

par parameter

t

par tangent?

\frac{x}{t} + t y = 2 c

x y = c^{2}

par

t

par normal?

x t^{3} - y t = c (t^{4} - 1)

x y = c^{2}

par

t_{1}, t_{2}

ko join karne wali chord?

x + t_{1} t_{2} y = c (t_{1} + t_{2})

t_{1}, t_{2}

par tangents ka intersection?

(\frac{2 c t _{1} t _{2}}{t _{1} + t _{2}}, \frac{2 c}{t _{1} + t _{2}})

x y = c^{2}

ki constant-area property?

Kisi bhi point se axes tak bane rectangle ka area

c^{2}

hota hai.

x y = c^{2}

ke vertices?

(c, c)

aur

(- c, - c)

line

y = x

par.

x^{2} - y^{2} = a^{2}

aur

x y = c^{2}

ka kya relation hai?

Axes ko

4 5^{\circ}

rotate karo; tab

c^{2} = a^{2} /2

Connections

Hyperbola standard form — $x y = c^{2}$ $a = b$ wala special case hai jo $4 5^{\circ}$ rotate hua hai.
Rotation of axes — woh tool jo $x^{2} - y^{2} = a^{2} \to x y = c^{2}$ convert karta hai.
Asymptotes of a hyperbola — yahan yeh coordinate axes ban jaate hain.
Tangent and Normal to conics — upar dikhaya gaya parametric method.
Eccentricity — equilateral hyperbolas ke liye $e = 2$ kyun hota hai.
Reciprocal function y=1/x — graph $y = c^{2} / x$ exactly yahi curve hai.