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CBSE · Class 11 · Mathematics

Ellipse (Ch 10: 10.5)

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

Ellipse: definition and parts

From daily life

Fix two pins, loop a string round them and draw with a pencil keeping the string tight: you draw an ellipse. Planets move on ellipses.

Simple definition

An ellipse is the set of points whose distances from two fixed points F₁ and F₂ (foci) add up to a constant 2a. The midpoint of F₁F₂ is the centre. The longest chord through the centre is the major axis (length 2a), the shortest is the minor axis (length 2b).

Formula: PF₁ + PF₂ = 2a

F₂F₁

In the figure: The ellipse x²/25 + y²/9 = 1 has major axis 10 and minor axis 6; the foci are (±4, 0).

Rule 1: Parts — Identify a, b and the axes.

  • Easy: x²/25 + y²/9 = 1: a = 5, b = 3.
  • Medium: Major axis length 2a = 10, minor axis 2b = 6.
  • Exam-type: Vertices (±5, 0); ends of minor axis (0, ±3).

Rule 2: String property — Sum of distances to the foci is constant.

  • Easy: If 2a = 10 then PF₁ + PF₂ = 10 for every point P.
  • Medium: A point at the end of the major axis: distances a − c and a + c add up to 2a.
  • Exam-type: If the foci coincide the ellipse becomes a circle.
Remember

The ellipse is longer along the major axis.

Common mistake

Thinking the foci are at the ends of the minor axis; they lie on the major axis.

Memory clue

Two pins and a string.

Practice questions

Easy
  1. 1The sum of focal distances of an ellipse is 14. Find a.
  2. 2Find the major axis of x²/16 + y²/9 = 1.
  3. 3Find the minor axis of x²/16 + y²/9 = 1.
Medium
  1. 4Where are the vertices of x²/25 + y²/16 = 1?
  2. 5A 20 cm string has its ends tied to two pins 12 cm apart. Find a, c and b of the ellipse traced when the string is kept taut.
  3. 6In an ellipse the focal distances of a point are 7 and 5. Find the length of the major axis.
Challenging
  1. 7What happens to an ellipse if both foci are at the centre?
  2. 8Find b² if a = 5 and c = 3.
Show answers (try them yourself first)
  1. 12a = 14, a = 7
  2. 22a = 8
  3. 32b = 6
  4. 4(±5, 0)
  5. 52a = 20, so a = 10; 2c = 12, so c = 6; b² = 100 − 36 = 64, so b = 8
  6. 612
  7. 7It becomes a circle
  8. 8b² = a² − c² = 16

Standard equation of an ellipse

From daily life

Put the centre at the origin and the major axis along the x-axis: the equation takes its simplest form.

Simple definition

With centre at the origin and major axis along the x-axis: x²/a² + y²/b² = 1 (a > b), foci (±c, 0), with c² = a² − b². If the major axis is along the y-axis: x²/b² + y²/a² = 1, foci (0, ±c).

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

Rule 1: Find foci and vertices — Larger denominator is a².

  • Easy: x²/25 + y²/9 = 1: a = 5, b = 3, c = 4: foci (±4, 0), vertices (±5, 0).
  • Medium: x²/9 + y²/25 = 1: major axis along y; a = 5, b = 3, c = 4: foci (0, ±4).
  • Exam-type: x²/13 + y²/4 = 1: c² = 9, foci (±3, 0).

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

  • Easy: a = 5, c = 3, major axis along x: b² = 16: x²/25 + y²/16 = 1.
  • Medium: Vertices (±4, 0), minor axis 2b = 6: x²/16 + y²/9 = 1.
  • Exam-type: Foci (0, ±2), a = 3: b² = 5: x²/5 + y²/9 = 1.
Remember

Always c² = a² − b² (the larger denominator minus the smaller).

Common mistake

Using c² = a² + b² (that is for the hyperbola).

Memory clue

Ellipse: a² minus b². Hyperbola: a² plus b².

Spot the difference

EllipseHyperbola
x²/a² + y²/b² = 1x²/a² − y²/b² = 1
c² = a² − b²c² = a² + b²
e < 1e > 1

Practice questions

Easy
  1. 1Find the foci of x²/16 + y²/7 = 1.
  2. 2Find the foci of x²/9 + y²/25 = 1.
  3. 3Find the vertices of x²/36 + y²/20 = 1.
Medium
  1. 4Find the equation with a = 5, b = 4 and major axis on the x-axis.
  2. 5Find the equation with foci (±3, 0) and a = 5.
  3. 6Find the equation of the ellipse with vertices (0, ±6) and foci (0, ±4).
Challenging
  1. 7Does (3, 0) lie on x²/9 + y²/4 = 1?
  2. 8Find the ellipse 4x² + 9y² = 36 in standard form and its foci.
Show answers (try them yourself first)
  1. 1c² = 9: (±3, 0)
  2. 2c² = 16: (0, ±4)
  3. 3(±6, 0)
  4. 4x²/25 + y²/16 = 1
  5. 5b² = 16: x²/25 + y²/16 = 1
  6. 6a = 6, c = 4, b² = 20: x²/20 + y²/36 = 1
  7. 79/9 = 1, yes
  8. 8x²/9 + y²/4 = 1; c² = 5, foci (±√5, 0)

Eccentricity and latus rectum of an ellipse

From daily life

Is the ellipse almost a circle or very flat? Eccentricity answers with one number between 0 and 1.

Simple definition

Eccentricity e = c/a (0 < e < 1). It is near 0 for a nearly circular ellipse. The latus rectum (chord through a focus perpendicular to the major axis) has length 2b²/a.

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

Rule 1: Find e — Find c first.

  • Easy: x²/25 + y²/9 = 1: c = 4, e = 4/5.
  • Medium: x²/16 + y²/7 = 1: c = 3, e = 3/4.
  • Exam-type: x²/9 + y²/9 = 1 (a circle): c = 0, e = 0.

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

  • Easy: x²/25 + y²/9 = 1: 2(9)/5 = 18/5.
  • Medium: x²/16 + y²/7 = 1: 2(7)/4 = 7/2.
  • Exam-type: x²/9 + y²/25 = 1 (major axis y): 2(9)/5 = 18/5.
Remember

For an ellipse e is always less than 1; closer to 1 means flatter.

Common mistake

Using the larger denominator for b².

Memory clue

e = c over a (focus distance over half-axis).

Practice questions

Easy
  1. 1Find e of x²/25 + y²/16 = 1.
  2. 2Find e of x²/100 + y²/36 = 1.
  3. 3Find the latus rectum of x²/36 + y²/20 = 1.
Medium
  1. 4Find e if a = 13 and b = 5.
  2. 5An ellipse has e = 3/5 and a = 10. Find b.
  3. 6Find the foci and e of 9x² + 16y² = 144.
Challenging
  1. 7Find the latus rectum of the ellipse with a = 6 and e = 1/2.
  2. 8Which ellipse is flatter: e = 0.3 or e = 0.9?
Show answers (try them yourself first)
  1. 1c = 3, e = 3/5
  2. 2c = 8, e = 4/5
  3. 32(20)/6 = 20/3
  4. 4c = 12, e = 12/13
  5. 5c = 6, b² = 64, b = 8
  6. 6x²/16 + y²/9 = 1; c² = 7; foci (±√7, 0); e = √7/4
  7. 7c = 3, b² = 27, latus rectum = 2(27)/6 = 9
  8. 8e = 0.9