Hyperbola — Complete JEE Revision Capsule
Compact, exam-focused hyperbola notes: standard & shifted forms, asymptotes, foci, conjugate axis, parametric forms (trig & hyperbolic), tangents & normals, chord formulae, latus rectum, and JEE traps — all MathJax-ready.
$\\dfrac{x^2}{a^2} - \\dfrac{y^2}{b^2} = 1$ (transverse axis along x)
$c^2 = a^2 + b^2,\ e=\\dfrac{c}{a}=\\sqrt{1+\\dfrac{b^2}{a^2}}$
$y=\\pm\\dfrac{b}{a}x$ (centered); combined: $\\dfrac{x^2}{a^2}-\\dfrac{y^2}{b^2}=0$
1. Definition & Standard Forms
Hyperbola is the locus of points for which the absolute difference of distances from two fixed points (foci) is constant: $$|PF_1 - PF_2| = 2a.$$
Standard centered forms:
- Transverse axis along x: $$\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$$
- Transverse axis along y: $$\dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1$$
Conjugate hyperbola (swap signs): $$\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = -1$$ (or equivalently $$\dfrac{y^2}{b^2} - \dfrac{x^2}{a^2} = 1$$ )
2. Key Quantities
- Foci: $(\pm c,0)$ with $$c^2 = a^2 + b^2.$$
- Eccentricity: $$e = \dfrac{c}{a} = \sqrt{1+\dfrac{b^2}{a^2}} \ (>1).$$
- Transverse axis length: $2a$; Conjugate axis length: $2b$ (geometric roles differ from ellipse).
- Latus rectum (through focus, perpendicular to transverse axis): length $$\ell = \dfrac{2b^2}{a}.$$
- Difference of distances to foci for any point on hyperbola = $2a$ (constant).
3. Asymptotes & Behavior at Infinity
Asymptotes of centered hyperbola $x^2/a^2 - y^2/b^2 = 1$ are straight lines:
$$ y = \pm \dfrac{b}{a} x. $$
Combined asymptote equation (by dropping constant term):
$$ \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 0. $$
Shifted center (h,k): asymptotes: $$ y - k = \pm \dfrac{b}{a}(x - h) $$ for hyperbola $\\dfrac{(x-h)^2}{a^2} - \\dfrac{(y-k)^2}{b^2} = 1$.
4. Parametric Forms
Two commonly used parameterizations (choose by context):
- Trigonometric (real parameter t): $$ x = a\sec t,\qquad y = b\tan t \quad (t\in\mathbb{R},\ \sec t\ge1). $$
- Hyperbolic (useful for continuous parameter u): $$ x = a\cosh u,\qquad y = b\sinh u \quad (u\in\mathbb{R}). $$
Derivative (slope) using param (trig): $$\dfrac{dy}{dx} = \dfrac{b\sec t\tan t}{a\sec t} = \dfrac{b}{a}\tan t.$$
5. Tangent Equations (Standard Forms)
Point form: tangent at $(x_1,y_1)$ on hyperbola:
$$ \boxed{\,\dfrac{x x_1}{a^2} - \dfrac{y y_1}{b^2} = 1\,} $$
Parametric tangent (using $t$):
For point $(a\sec t, b\tan t)$: $$ \boxed{\,\dfrac{x\sec t}{a} - \dfrac{y\tan t}{b} = 1\,}. $$
Slope form: For line $y=mx+c$ to be tangent, discriminant = 0 ⇒
$$ \boxed{\,c^2 = a^2 m^2 - b^2\,}. $$ So tangents with slope $m$ are $$ y = mx \pm \sqrt{a^2 m^2 - b^2}\,, $$ provided $a^2 m^2 \ge b^2$.
6. Normals & Polars
Normal at point $(x_1,y_1)$: one convenient form (algebraic):
$$ \boxed{\,\dfrac{a^2 x}{x_1} - \dfrac{b^2 y}{y_1} = a^2 - b^2\,} $$ (Can be rearranged to point-slope form using slope $m_n = -\dfrac{a^2 y_1}{b^2 x_1}$ appropriately.)
Polar (T=0): For hyperbola S: $x^2/a^2 - y^2/b^2 -1=0$, the polar of $(x_1,y_1)$ is $$\dfrac{x x_1}{a^2} - \dfrac{y y_1}{b^2} = 1\ (\,\text{same as tangent if }(x_1,y_1)\text{ lies on curve}\,).$$
7. Chords & Focal Chords
Chord joining parameter values $t_1,t_2$ (trig param) can be written via elimination of parameters. For focal chords (through focus) parameters satisfy special relations analogous to ellipse.
Focal chord: if a chord passes through focus $(c,0)$ then parameters satisfy relation (use param forms & eliminate) — many JEE problems ask specific numeric cases; treat algebraically using param coordinates.
Latus rectum: through focus; its half-length (measured parallel to conjugate axis) = $\\dfrac{b^2}{a}$ so full length $$\\ell = \\dfrac{2b^2}{a}.$$
8. Conjugate Hyperbola & Relations
Conjugate hyperbola of $\\dfrac{x^2}{a^2} - \\dfrac{y^2}{b^2} = 1$ is $\\dfrac{y^2}{b^2} - \\dfrac{x^2}{a^2} = 1$. Their asymptotes are same; foci swap roles between transverse & conjugate axes.
Product of equations of hyperbola and its conjugate gives central rectangular hyperbola of second degree (useful in some problems).
9. Region & Point Condition
Define function for point $P(x_1,y_1)$:
$$ S = \\dfrac{x_1^2}{a^2} - \\dfrac{y_1^2}{b^2} - 1. $$
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10. Useful Identities & Quick Tricks
- Difference of distances to foci = $2a$ for any point on hyperbola.
- Equation of asymptotes (shifted center): find by linearizing highest degree terms.
- To check tangency of $y=mx+c$: verify $c^2 = a^2 m^2 - b^2$ and $a^2 m^2 \ge b^2$.
- For rotated hyperbolas ($Bxy$ term present), rotate axes using $\\tan 2\\theta = \\dfrac{B}{A-C}$.
11. Worked Examples (High-yield)
Example 1: For hyperbola $\\dfrac{x^2}{9} - \\dfrac{y^2}{4} = 1$, find foci, asymptotes and eccentricity.
Solution: $a^2=9\\Rightarrow a=3$, $b^2=4\\Rightarrow b=2$. $c^2=a^2+b^2=9+4=13\\Rightarrow c=\\sqrt{13}$. $e=c/a=\\dfrac{\\sqrt{13}}{3}$. Asymptotes: $y=\\pm\\dfrac{b}{a}x=\\pm\\dfrac{2}{3}x$.
Example 2: Tangent(s) with slope $m=1$ to hyperbola $\\dfrac{x^2}{16}-\\dfrac{y^2}{9}=1$.
Solution: Use $c^2=a^2m^2-b^2 =16\\cdot1 - 9 =7$. So tangents: $y=x\\pm\\sqrt7$ (valid since $16\\cdot1 \\ge 9$).
Example 3: Find equations of asymptotes for shifted hyperbola $\\dfrac{(x-2)^2}{4} - \\dfrac{(y+1)^2}{1} = 1$.
Solution: Asymptotes: $y+1 = \\pm \\dfrac{1}{2}(x-2)$ → or $y = -1 \\pm \\dfrac{1}{2}(x-2)$.
12. JEE Traps & How to Avoid
- Trap: mixing ellipse sign rules — hyperbola has minus between x^2 and y^2; check which term is positive to know transverse axis.
- Trap: forget asymptote condition $a^2 m^2 \\ge b^2$ for real tangents with slope $m$.
- Trap: sign of S for point location — memorize interpretation for branches vs interior between branches.
- Tip: use parametric forms for tangents & normals; hyperbolic param often simplifies integrals or continuous variants.
13. Practice Problems (Try first)
- Find equation of tangent at point corresponding to $t=\\pi/4$ (trig param) on $\\dfrac{x^2}{a^2}-\\dfrac{y^2}{b^2}=1$.
- Show that asymptotes of $\\dfrac{x^2}{9}-\\dfrac{y^2}{4}=1$ are $y=\\pm\\tfrac{2}{3}x$ and verify intersection at origin.
- Find latus rectum length for $\\dfrac{x^2}{a^2}-\\dfrac{y^2}{b^2}=1$.
- For hyperbola $\\dfrac{x^2}{4}-\\dfrac{y^2}{1}=1$, find tangent(s) parallel to $y=3x+1$.
- Find equation of hyperbola with foci at $(\\pm5,0)$ and transverse axis length $8$.
Show Answers & Hints
- Parametric point: $(a\\sec t, b\\tan t)$; substitute into point form tangent.
- Compute as in worked example 1.
- Use formula $\\ell=\\dfrac{2b^2}{a}$.
- Solve condition $c^2 = a^2 m^2 - b^2$ with $m=3$; check feasibility.
- Given foci ±5 ⇒ c=5, transverse 2a=8 ⇒ a=4 ⇒ b^2=c^2-a^2=25-16=9 ⇒ hyperbola: $x^2/16 - y^2/9 = 1$.
14. One-line Takeaways
- Sign between squared terms tells which axis is transverse (positive term's variable gives transverse axis direction).
- Asymptotes are your first tool — linearize the highest-degree part for quick answers.
- Parametric forms (sec–tan or cosh–sinh) speed up tangent/normal problems.
- Remember $c^2=a^2+b^2$ and latus rectum $=2b^2/a$ — these recur in JEE.
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