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CBSE · Class 11 · Mathematics

Hyperbola (Ch 10: 10.6)

3 concepts · 24 practice questions with answers · Free to read

Hyperbola: definition and parts

From daily life

Two cooling towers or the sound of a sonic boom on the ground: curves with two separate branches. They are hyperbolas.

Simple definition

A hyperbola is the set of points whose distances from two fixed points F₁, F₂ (foci) differ by a constant 2a. It has two branches. The line through the foci is the transverse axis (length 2a); the points where it meets the curve are the vertices.

Formula: |PF₁ − PF₂| = 2a

F₂F₁

In the figure: The hyperbola x²/9 − y²/4 = 1: vertices (±3, 0), foci (±√13, 0).

Rule 1: Parts — a, b and c.

  • Easy: x²/9 − y²/4 = 1: a = 3, b = 2, c = √13.
  • Medium: Vertices (±3, 0), foci (±√13, 0).
  • Exam-type: The two branches open left and right.

Rule 2: Difference property — The difference of focal distances is 2a.

  • Easy: If 2a = 6 then |PF₁ − PF₂| = 6 for every point.
  • Medium: A vertex is at distance c − a and c + a from the foci: difference 2a.
  • Exam-type: Compare: ellipse = sum constant; hyperbola = difference constant.
Remember

The positive term (x² or y²) shows along which axis the branches open.

Common mistake

Thinking a hyperbola has one branch.

Memory clue

Hyperbola = Difference. Ellipse = Sum.

Practice questions

Easy
  1. 1The difference of focal distances of a hyperbola is 8. Find a.
  2. 2How many branches does a hyperbola have?
  3. 3Find a and b of x²/16 − y²/9 = 1.
Medium
  1. 4Find the vertices of x²/25 − y²/16 = 1.
  2. 5Which axis do the branches of y²/9 − x²/16 = 1 open along?
  3. 6A point is 15 from F₁ and 9 from F₂. Find the length of the transverse axis.
Challenging
  1. 7Find c if a = 5 and b = 12.
  2. 8Find a if the vertices are (±7, 0).
Show answers (try them yourself first)
  1. 14
  2. 22
  3. 3a = 4, b = 3
  4. 4(±5, 0)
  5. 5The y-axis
  6. 62a = 15 − 9 = 6
  7. 7c = 13
  8. 87

Standard equation of a hyperbola

From daily life

With the centre at the origin and foci on the x-axis the equation of a hyperbola is neat.

Simple definition

With foci on the x-axis: x²/a² − y²/b² = 1, foci (±c, 0), vertices (±a, 0), with c² = a² + b². With foci on the y-axis: y²/a² − x²/b² = 1, foci (0, ±c).

Formula: x²/a² − y²/b² = 1, c² = a² + b²

Rule 1: Find foci and vertices — The positive term gives the axis; a² is its denominator.

  • Easy: x²/16 − y²/9 = 1: c = 5: foci (±5, 0), vertices (±4, 0).
  • Medium: y²/9 − x²/16 = 1: foci (0, ±5), vertices (0, ±3).
  • Exam-type: x²/4 − y²/5 = 1: c = 3, foci (±3, 0).

Rule 2: Write the equation — Find a² and b².

  • Easy: a = 3, c = 5, foci on x-axis: b² = 16: x²/9 − y²/16 = 1.
  • Medium: Vertices (0, ±4), foci (0, ±5): b² = 9: y²/16 − x²/9 = 1.
  • Exam-type: a = 2, b = 3: x²/4 − y²/9 = 1.
Remember

In a hyperbola a need not be larger than b.

Common mistake

Using c² = a² − b² (that is for the ellipse).

Memory clue

Hyperbola: a² plus b² makes c².

Practice questions

Easy
  1. 1Find the foci of x²/9 − y²/16 = 1.
  2. 2Find the foci of y²/25 − x²/144 = 1.
  3. 3Find the vertices of x²/49 − y²/25 = 1.
Medium
  1. 4Write the hyperbola with a = 4, b = 3, foci on the x-axis.
  2. 5Write the hyperbola with vertices (±3, 0) and foci (±5, 0).
  3. 6Write the hyperbola with vertices (0, ±5) and foci (0, ±13).
Challenging
  1. 7Convert 16x² − 9y² = 144 to standard form.
  2. 8Does (5, 0) lie on x²/25 − y²/9 = 1?
Show answers (try them yourself first)
  1. 1(±5, 0)
  2. 2c = 13: (0, ±13)
  3. 3(±7, 0)
  4. 4x²/16 − y²/9 = 1
  5. 5b² = 16: x²/9 − y²/16 = 1
  6. 6b² = 144: y²/25 − x²/144 = 1
  7. 7x²/9 − y²/16 = 1
  8. 8Yes (a vertex).

Eccentricity and latus rectum of a hyperbola

From daily life

How "open" are the branches? Eccentricity greater than 1 tells us.

Simple definition

Eccentricity e = c/a > 1. The latus rectum (chord through a focus perpendicular to the transverse axis) has length 2b²/a.

Formula: e = c/a, latus rectum = 2b²/a

Rule 1: Find e — Compute c from a² + b².

  • Easy: x²/9 − y²/16 = 1: c = 5, e = 5/3.
  • Medium: x²/16 − y²/9 = 1: c = 5, e = 5/4.
  • Exam-type: x²/1 − y²/1 = 1: c = √2, e = √2.

Rule 2: Find the latus rectum — Use 2b²/a.

  • Easy: x²/9 − y²/16 = 1: 2(16)/3 = 32/3.
  • Medium: x²/16 − y²/9 = 1: 2(9)/4 = 9/2.
  • Exam-type: y²/4 − x²/9 = 1: a = 2, b = 3: 2(9)/2 = 9.
Remember

e > 1 for a hyperbola, e = 1 for a parabola, e < 1 for an ellipse.

Common mistake

Using the b² for ellipse formula c² = a² − b².

Memory clue

Hyperbola: e is big (more than 1).

Practice questions

Easy
  1. 1Find e of x²/9 − y²/16 = 1.
  2. 2Find e of x²/16 − y²/9 = 1.
  3. 3Find the latus rectum of x²/25 − y²/144 = 1.
Medium
  1. 4Find e if a = 3 and c = 5.
  2. 5A hyperbola has e = 5/4 and a = 4. Find b.
  3. 6Find the foci and e of 9x² − 16y² = 144.
Challenging
  1. 7Find the latus rectum of the hyperbola with a = 4, e = 5/4.
  2. 8Order e for circle, ellipse, parabola, hyperbola.
Show answers (try them yourself first)
  1. 15/3
  2. 25/4
  3. 32(144)/5 = 288/5
  4. 45/3
  5. 5c = 5, b = 3
  6. 6x²/16 − y²/9 = 1; c = 5; foci (±5, 0); e = 5/4
  7. 7c = 5, b² = 9: 2(9)/4 = 9/2
  8. 80, between 0 and 1, 1, greater than 1