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Hyperbola

Part of the CUET UG study roadmap. Mathematics topic math-007 of Mathematics.

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Hyperbola

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

A hyperbola is the locus of points in a plane whose distances from two fixed points (the foci) have a constant difference equal to 2a. It belongs to the family of conic sections and is distinguished by eccentricity e > 1.

  • Horizontal standard form: x²/a² − y²/b² = 1, with foci at (±ae, 0) and vertices at (±a, 0).
  • Key relation: b² = a²(e² − 1); latus rectum length = 2b²/a; asymptotes y = ±(b/a)x.
  • Quick identifier: the term with the positive sign sits on the transverse axis.

Exam pointer: For CUET UG, expect one MCQ asking you to find eccentricity from a given a and b, or to match a hyperbola equation with its foci. Memorise b² = a²(e² − 1), not the ellipse version.

🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Definition and Orientation

Geometrically, a hyperbola is the set of points P such that |PS − PS′| = 2a, where S and S′ are the foci. This constant difference (not sum) separates it from the ellipse. The standard equation comes in two orientations, depending on which squared term is positive.

Standard Equations and Key Elements

For x²/a² − y²/b² = 1, the centre is (0, 0), the transverse axis lies along the x-axis, and the two branches open left and right. For y²/a² − x²/b² = 1, the branches open upward and downward.

ElementHorizontal form x²/a² − y²/b² = 1Vertical form y²/a² − x²/b² = 1
Centre(0, 0)(0, 0)
Vertices(±a, 0)(0, ±a)
Foci(±ae, 0)(0, ±ae)
Transverse axis length2a2a
Conjugate axis length2b2b
Asymptotesy = ±(b/a)xy = ±(a/b)x

Eccentricity and the b² Relation

Because e > 1 always, the relation linking a, b, e is b² = a²(e² − 1), so e = √(1 + b²/a²). A common trap is writing b² = a²(1 − e²), which only holds for ellipses.

Parametric Form

Any point on x²/a² − y²/b² = 1 can be written as (a sec θ, b tan θ), where θ ∈ [−π/2, π/2] ∪ [π/2, 3π/2] traces one branch.

  • Rectangular hyperbola: occurs when a = b, giving x² − y² = a², eccentricity √2, and perpendicular asymptotes.
  • Focal chord check: the difference of focal distances from any point on the curve equals 2a, useful in assertion-reason MCQs.
  • CUET pattern: most questions are direct formula applications — compute e from given a, b; find foci; or identify asymptotes from an equation.

🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Worked Example

Find the eccentricity, foci, and latus rectum length of 9x² − 16y² = 144.

Divide by 144: x²/16 − y²/9 = 1, so a² = 16, b² = 9, a = 4, b = 3. Eccentricity: e = √(1 + 9/16) = √(25/16) = 5/4. Foci: (±ae, 0) = (±4 × 5/4, 0) = (±5, 0). Latus rectum: 2b²/a = 2(9)/4 = 9/2 = 4.5 units.

Common Mistakes and Edge Cases

MistakeWhy it failsCorrect approach
Using sum = 2aThat defines an ellipseHyperbola uses |difference| = 2a
Foci written as (±a, 0)Forgot eccentricity factorFoci are (±ae, 0)
b² = a²(1 − e²)Ellipse relationHyperbola: b² = a²(e² − 1)
Dropping sign on swapLoses orientation infoKeep positive term aligned with transverse axis
e < 1 claimedViolates definitione must satisfy e > 1

Connections and Strategy

Hyperbola ties directly to Parabola and Ellipse under the Conic Sections unit, and the eccentricity classification (e = 0 circle, 0 < e < 1 ellipse, e = 1 parabola, e > 1 hyperbola) is a recurring CUET MCQ. The focal property (constant difference) and the parametric form are also asked in assertion-reason items.

Practice prompts:

  1. For 16x² − 25y² = 400, find a, b, eccentricity, foci, vertices, and the latus rectum length.
  2. The hyperbola passes through (5, 2) and (√29, 9/4) with centre (0, 0) and transverse axis along x-axis. Determine its standard equation.

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