Hyperbola — standard forms, asymptotes, foci, eccentricity

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1. Deriving the standard equation from scratch

Setup (WHY this coordinate choice): put the foci symmetrically on the $x$-axis at $F_1(-c,0)$ and $F_2(c,0)$ so the algebra is symmetric and the centre sits at the origin.

Let $P=(x,y)$. Then $\sqrt{(x+c)^2+y^2}-\sqrt{(x-c)^2+y^2}=\pm 2a.$

Step 1 — isolate one root. Why? One square root at a time is manageable. $\sqrt{(x+c)^2+y^2}=\pm 2a+\sqrt{(x-c)^2+y^2}.$

Step 2 — square both sides. Why? Kills the isolated radical. $(x+c)^2+y^2 = 4a^2 \pm 4a\sqrt{(x-c)^2+y^2}+(x-c)^2+y^2.$

Expand and cancel $x^2, c^2, y^2$: $4cx = 4a^2 \pm 4a\sqrt{(x-c)^2+y^2}\;\Rightarrow\; cx-a^2=\pm a\sqrt{(x-c)^2+y^2}.$

Step 3 — square again. Why? Remove the last radical. $c^2x^2 -2a^2cx+a^4 = a^2\big[(x-c)^2+y^2\big]=a^2x^2-2a^2cx+a^2c^2+a^2y^2.$

Cancel $-2a^2cx$, regroup: $(c^2-a^2)x^2 - a^2y^2 = a^2(c^2-a^2).$

Step 4 — name the leftover. Since $c>a$, the quantity $c^2-a^2>0$. Define $\boxed{b^2 = c^2 - a^2}.$ Why define $b$? It makes the equation clean AND turns out to be the vertical semi-extent that fixes the asymptote slopes. Divide by $a^2b^2$:

The vertical version just swaps roles: $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1 \quad(\text{foci on the }y\text{-axis},\ (0,\pm c)).$

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2. Eccentricity — the shape number

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

Directrix form: each focus pairs with a directrix line $x=\pm a/e$, and for every point $P$: $\frac{PF}{\text{(distance to that directrix)}}=e.$ This is the single unifying definition of all conics.

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3. Asymptotes — deriving the straight-line guides

Derivation. Solve for $y$: $y^2=b^2\left(\frac{x^2}{a^2}-1\right)\;\Rightarrow\; y=\pm\frac{b}{a}\,x\sqrt{1-\frac{a^2}{x^2}}.$ As $|x|\to\infty$, $\sqrt{1-a^2/x^2}\to 1$, so $y\to \pm\frac{b}{a}x.$

Gap → 0 proof (Why they never cross): the vertical gap between curve and line is $\frac{b}{a}x-\frac{b}{a}x\sqrt{1-\tfrac{a^2}{x^2}}=\frac{b}{a}x\left(1-\sqrt{1-\tfrac{a^2}{x^2}}\right)\xrightarrow{x\to\infty}0,$ because the bracket shrinks like $\frac{a^2}{2x^2}$ and $x$ only grows linearly. Product $\to 0$: the curve hugs but never meets the line.

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4. Rectangular (equilateral) hyperbola

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Worked Examples

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Flashcards

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Recall Feynman: explain to a 12-year-old

Imagine two thumbtacks (the foci) and you walk so that one tack is always exactly, say, 4 steps closer than the other. Not the total — the difference. If you keep that difference fixed and trace where you can stand, you get a curve that swoops toward one tack, then a mirror copy swoops toward the other: two separate arcs. Way out at the edges, these arcs straighten out and run almost perfectly alongside two crossing straight rulers (the asymptotes) — getting closer and closer forever but never quite touching, like a car racing beside a wall it never scrapes. How "wide open" the arcs are is one number, $e$; for a hyperbola $e$ is always bigger than 1.

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Connections

• Ellipse — standard forms, foci, eccentricity — same derivation with $+ \to -$ and $c^2=a^2-b^2$.
• Parabola — standard forms, focus, directrix — the borderline case $e=1$.
• Eccentricity — unified conic definition — one focus–directrix ratio classifies all conics.
• Asymptotes and Limits at Infinity — the gap-goes-to-zero argument is a limit.
• Rectangular Hyperbola xy=c^2 — $45^\circ$ rotation of $a=b$ case.
• Conic Sections from a Double Cone — steep cut through both nappes.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, hyperbola samajhne ka sabse aasan tareeka yeh hai: ellipse mein hum do foci se distances ko add karte the aur constant rakhte the. Hyperbola mein bas ek operation badal jaata hai — ab hum distances ka difference constant rakhte hain, yaani $|PF_1 - PF_2| = 2a$. Yeh chhota sa minus sign hi wajah hai ki curve do alag-alag branches mein tut jaata hai jo infinity ki taraf bhaagti hain, band nahi hoti.

Standard equation $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ hai. Yaad rakho sabse important cheez: hyperbola mein foci vertices ke bahar hote hain, isliye $c>a$ aur relation $c^2 = a^2 + b^2$ (plus!) — ellipse waala minus yahan mat lagana, yeh sabse common galti hai. Eccentricity $e=c/a$ hamesha 1 se bada hota hai. Aur ek shortcut: $e=\sqrt{1+b^2/a^2}$ — isse tumhe $a,b$ alag se nahi chahiye.

Asymptotes woh do straight lines hain $y=\pm\frac{b}{a}x$ jinke paas curve door jaakar chipak jaata hai par kabhi touch nahi karta. Kaise nikaale? Equation mein jab $x$ bahut bada ho, toh "$1$" ignore ho jaata hai, aur bacha $\frac{x^2}{a^2}-\frac{y^2}{b^2}=0$ — yeh do lines hi hain. Slope $b/a$ hota hai (na ki $a/b$), yeh central box ke corner $(a,b)$ se check kar lo.

Yeh cheez real life mein bhi kaam aati hai — jaise navigation systems mein jahan do stations se signal ka time-difference constant hota hai, wahan tumhari position ek hyperbola pe aati hai. Toh concept clear rakho: difference constant, plus sign, $e>1$, slope $b/a$. Bas!