The ellipse — CSCA (China)

In this lesson you will learn

Use the definition of an ellipse and write its standard equation from the given data
Read a, b, c, the foci, the vertices and the eccentricity from an equation
Compute the perimeters and areas of focal triangles and the focal radii
Decide how a line meets an ellipse with the discriminant and find chord lengths and midpoints

Tie the two ends of a piece of string to two pins in a board, pull the string tight with a pencil and move the pencil around. The sum of its distances to the two pins always equals the length of the string, and the curve it draws is an ellipse. Gardeners mark oval flower beds this way, and the planets move along ellipses with the Sun at one of the two special points, the foci. In the CSCA syllabus the ellipse belongs to “Plane analytic geometry”. It builds on the lesson “Lines and circles in the coordinate plane” — a circle is an ellipse whose two foci have merged into the centre — and it comes before “The hyperbola” and “The parabola”. Here you will learn the definition, both standard equations, the numbers a, b, c and e, the triangles formed with the foci, and what happens when a line meets an ellipse.

Definition and standard equations

Definition

Ellipse (椭圆)

The set of all points P of the plane for which the sum of the distances to two fixed points F₁ and F₂, the foci (焦点), is a constant 2a greater than the distance between the foci: |PF₁| + |PF₂| = 2a > |F₁F₂| = 2c. The distance 2c is called the focal distance (焦距).

The condition 2a > 2c matters. If 2a = 2c, the only points with |PF₁| + |PF₂| = |F₁F₂| are the points of the segment F₁F₂. If 2a < 2c, there are no such points at all, by the triangle inequality. CSCA likes to test exactly this.

Where does the equation come from? Put the foci on the x-axis, symmetric about the origin: F₁(−c, 0), F₂(c, 0). The definition reads √((x + c)² + y²) + √((x − c)² + y²) = 2a. Move one root to the right side, square, simplify, and square again: (a² − c²)x² + a²y² = a²(a² − c²). Since a > c, the number a² − c² is positive; call it b² and divide by a²b².

x²/a² + y²/b² = 1 y²/a² + x²/b² = 1 (a > b > 0, b² = a² − c²)x²/a² + y²/b² = 1 y²/a² + x²/b² = 1 (a > b > 0, b² = a² − c²)

where:

asemi-major axis (half of the major axis)
bsemi-minor axis
chalf of the focal distance: foci (±c, 0) in the first equation and (0, ±c) in the second

The foci lie on the axis of the larger denominator: x²/16 + y²/25 = 1 has 25 under y², so its foci are on the y-axis. In Chinese textbooks a is always the larger number (a > b > 0), whichever variable it stands under.

Example 1: equations and foci

1) The foci are (−3, 0) and (3, 0), and the sum of the distances is 10. Write the equation.
2) Where are the foci of x²/7 + y²/16 = 1?
3) An ellipse with its foci on the y-axis has c = 2 and passes through (0, 3). Write its equation.
4) For which m is x²/(m − 2) + y²/(6 − m) = 1 an ellipse with its foci on the x-axis?

Show solution

1) 2a = 10, a = 5, c = 3, b² = 25 − 9 = 16: x²/25 + y²/16 = 1.
2) The larger denominator 16 stands under y², so the foci are on the y-axis: c² = 16 − 7 = 9, foci (0, ±3).
3) (0, 3) lies on the axis of the foci, so it is a vertex: a = 3, b² = 9 − 4 = 5: y²/9 + x²/5 = 1.
4) Both denominators must be positive and the one under x² larger: m − 2 > 6 − m > 0, so 4 < m < 6.

Example 2 (CSCA style): a sum of distances

F₁(−3, 0) and F₂(3, 0) are fixed, and a point P moves so that |PF₁| + |PF₂| = 6. What is the path of P?
A) an ellipse B) the segment F₁F₂ C) a circle D) there is no such point

Show solution

Here 2a = 6 equals |F₁F₂| = 6, so the condition 2a > 2c fails. For any P off the segment, |PF₁| + |PF₂| > |F₁F₂| by the triangle inequality, while every point of the segment gives exactly 6. The path is the segment F₁F₂, answer B.
A is the automatic answer of anyone who does not compare 2a with 2c. D would be right if the sum were less than 6.

a, b, c: vertices, axes and eccentricity

Definition

Vertices, major and minor axis (顶点, 长轴, 短轴)

The four points where the ellipse meets its axes of symmetry are its vertices. The segment A₁A₂ through the foci is the major axis (length 2a); the perpendicular segment B₁B₂ is the minor axis (length 2b). They cross at the centre of the ellipse.

Feature	x²/a² + y²/b² = 1	y²/a² + x²/b² = 1
Foci	(±c, 0)	(0, ±c)
Vertices	(±a, 0), (0, ±b)	(0, ±a), (±b, 0)
Major axis	on the x-axis, length 2a	on the y-axis, length 2a
Minor axis	on the y-axis, length 2b	on the x-axis, length 2b
Range	\|x\| ≤ a, \|y\| ≤ b	\|x\| ≤ b, \|y\| ≤ a

Both ellipses are symmetric about both axes and about the origin, their centre; the focal distance is 2c in both.

a² = b² + c²

where:

asemi-major axis: the distance from an end of the minor axis to a focus
bsemi-minor axis
cdistance from the centre to a focus

The end B₂(0, b) of the minor axis is equally far from both foci, and the two distances add up to 2a, so |B₂F₁| = |B₂F₂| = a. The right triangle OB₂F₂ has legs b and c and hypotenuse a: in an ellipse a is the largest of the three numbers.

The ellipse x²/a² + y²/b² = 1 with a = 5, b = 3, c = 4, drawn to scale: in the triangle OB₂F₂, a is the hypotenuse.

Example 3: reading a, b and c

1) Find the vertices, the foci and the lengths of the axes of x²/25 + y²/9 = 1.
2) Write 4x² + 9y² = 36 in standard form and find its foci.
3) The minor axis is 8 long and the focal distance is 6. Write the equation.

Show solution

1) a = 5, b = 3, c = √(25 − 9) = 4: vertices (±5, 0), (0, ±3), foci (±4, 0), major axis 10, minor axis 6, focal distance 8.
2) Divide by 36: x²/9 + y²/4 = 1; c² = 9 − 4 = 5, foci (±√5, 0).
3) b = 4, c = 3, a = 5. The position of the foci is not given, so there are two answers: x²/25 + y²/16 = 1 or y²/25 + x²/16 = 1.

e = c/a = √(1 − b²/a²), 0 < e < 1e = c/a = √(1 − b²/a²), 0 < e < 1

where:

eeccentricity (离心率)
a, b, cas above

e measures how flat the ellipse is. As e → 0, c → 0 and b → a: the ellipse becomes a circle. As e → 1, b → 0 and it flattens towards a segment. The ratio of the axes is b/a = √(1 − e²).

Example 4: eccentricity

1) Find e for x²/16 + y²/12 = 1.
2) The semi-minor axis equals half the focal distance (b = c). Find e.
3) The foci divide the major axis A₁A₂ into three equal parts. Find e.

Show solution

1) c² = 16 − 12 = 4, c = 2, e = 2/4 = 1/2.
2) a² = b² + c² = 2c², so a = √2·c and e = c/a = √2/2.
3) |A₁F₁| = |F₁F₂| = |F₂A₂| = 2a/3, so 2c = 2a/3 and e = 1/3.

Example 5 (CSCA style): an eccentricity with a parameter

The ellipse x²/9 + y²/m = 1 has eccentricity √3/2. What is m?
A) 9/4 B) 36 C) 9/4 or 36 D) 27/4

Show solution

The major axis is not given, so try both cases.
If m < 9: a² = 9, c² = 9 − m, and e² = (9 − m)/9 = 3/4 gives m = 9/4.
If m > 9: a² = m, c² = m − 9, and e² = (m − 9)/m = 3/4 gives m = 36. Answer C.
A and B each keep only one case. D puts b² instead of c² into e² = c²/a²: m/9 = 3/4.

Interactive

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Here a is the semi-axis along x and b the semi-axis along y. Keep a > b: e = √(1 − b²/a²) grows as the ellipse gets flatter, and a = b gives a circle. The line y = kx + m touches the ellipse when m² = a²k² + b² (see the shortcut below): for a = 4, b = 2 and k = 0.5 that is m = ±2√2 ≈ ±2.83.

Focal triangles

Join a point P of the ellipse to both foci: the triangle PF₁F₂ is called a focal triangle (焦点三角形). Two of its sides are tied together by the definition, |PF₁| + |PF₂| = 2a, and the third is fixed, |F₁F₂| = 2c. Almost every question about it is solved with these two facts and the law of cosines from the lesson “Solving triangles: the law of sines and the law of cosines”.

perimeter of △PF₁F₂ = 2a + 2c perimeter of △ABF₂ = 4a (chord AB through F₁)

where:

Pany point of the ellipse (not an end of the major axis)
ABa chord through the focus F₁

For the chord: |AF₁| + |AF₂| = 2a and |BF₁| + |BF₂| = 2a, while |AF₁| + |BF₁| = |AB|. Adding gives |AB| + |AF₂| + |BF₂| = 4a, whatever the direction of the chord.

Example 6: perimeters and distances to the foci

1) Find the perimeter of △PF₁F₂ for the ellipse x²/25 + y²/16 = 1.
2) A chord AB of x²/4 + y²/3 = 1 passes through F₁. Find the perimeter of △ABF₂.
3) P lies on x²/16 + y²/12 = 1 and |PF₁| = 3|PF₂|. Find |PF₁| and |PF₂|.

Show solution

1) a = 5, c = 3: 10 + 6 = 16.
2) a = 2, so the perimeter is 4a = 8, whatever the direction of the chord.
3) |PF₁| + |PF₂| = 2a = 8 and |PF₁| = 3|PF₂|, so |PF₂| = 2 and |PF₁| = 6. Since a − c = 4 − 2 = 2 is the smallest possible distance to a focus (see below), P is the right vertex (4, 0).

a − c ≤ |PF₁| ≤ a + c |PF₁| = a + ex₀, |PF₂| = a − ex₀

where:

P(x₀, y₀)a point of the ellipse x²/a² + y²/b² = 1
F₁(−c, 0), F₂(c, 0)left and right foci
eeccentricity

The distance from a point of the ellipse to a focus is called a focal radius (焦半径). Substituting y₀² = b²(1 − x₀²/a²) into |PF₂|² = (x₀ − c)² + y₀² gives (a − ex₀)², so the focal radius is linear in x₀: it is smallest, a − c, at the vertex nearest to the focus and largest, a + c, at the far vertex.

Example 7: focal radii

1) P lies on x²/25 + y²/9 = 1 and its x-coordinate is 5/2. Find |PF₁| and |PF₂|.
2) What are the greatest and the least distances from a point of x²/49 + y²/40 = 1 to a focus?

Show solution

1) c = 4, e = 4/5: |PF₁| = 5 + (4/5)(5/2) = 7, |PF₂| = 5 − 2 = 3. Check: 7 + 3 = 10 = 2a ✓.
2) c² = 49 − 40 = 9, c = 3: greatest a + c = 10, least a − c = 4, at the two ends of the major axis.

S = b² · tan(θ/2), θ = ∠F₁PF₂S = b² · tan(θ/2), θ = ∠F₁PF₂

where:

Sarea of triangle PF₁F₂
θthe angle at P
bsemi-minor axis

Derivation: let m = |PF₁| and n = |PF₂|. The law of cosines gives 4c² = m² + n² − 2mn cos θ = (m + n)² − 2mn(1 + cos θ) = 4a² − 2mn(1 + cos θ), so mn = 2b²/(1 + cos θ). Then S = ½mn sin θ = b² sin θ/(1 + cos θ) = b² tan(θ/2). For θ = 90°: S = b².

Example 8 (CSCA style): the area of a focal triangle

P is a point of the ellipse x²/9 + y²/4 = 1 with foci F₁ and F₂, and ∠F₁PF₂ = 60°. What is the area of triangle F₁PF₂?
A) 4√3/3 B) 4√3 C) 2√3 D) 4

Show solution

S = b² tan(θ/2) = 4 · tan 30° = 4 · √3/3 = 4√3/3, answer A.
B uses tan 60° instead of tan 30°; D is b², the area for a right angle. Check without the formula: m + n = 6 and 4c² = 20 = (m + n)² − 3mn = 36 − 3mn, so mn = 16/3 and S = ½ · (16/3) · (√3/2) = 4√3/3 ✓.

Example 9: a right angle at P

1) P lies on x²/49 + y²/24 = 1 and PF₁ ⊥ PF₂. Find the area of △F₁PF₂ and the lengths |PF₁| and |PF₂|.
2) Does the ellipse x²/9 + y²/5 = 1 have a point P with PF₁ ⊥ PF₂?

Show solution

1) a = 7, b² = 24, c = 5. S = b² = 24. Also mn = 2b² = 48 and m + n = 14, so m and n are the roots of t² − 14t + 48 = 0: 8 and 6 — the 6-8-10 right triangle with hypotenuse 2c = 10.
2) The points P with ∠F₁PF₂ = 90° lie on the circle with diameter F₁F₂, whose radius is c. Here c = 2 < b = √5, so this circle stays inside the ellipse: there is no such point. The largest angle F₁PF₂, reached at an end of the minor axis, is less than 90°.

A line and an ellipse

A point P(x₀, y₀) lies inside the ellipse x²/a² + y²/b² = 1 if x₀²/a² + y₀²/b² < 1, on it if the sum equals 1, and outside if it is greater than 1. A line meets an ellipse in at most two points. Which case occurs is decided by a quadratic equation, because the distance trick of the circle does not work for an ellipse.

1
Substitute
Put y = kx + m into the equation of the ellipse and clear the fractions: (b² + a²k²)x² + 2a²kmx + a²(m² − b²) = 0.
2
Discriminant
Δ > 0: two common points (a secant); Δ = 0: one common point (a tangent); Δ < 0: no common points.
3
Vieta’s formulas
For a secant, x₁ + x₂ = −2a²km/(b² + a²k²) and x₁x₂ = a²(m² − b²)/(b² + a²k²); the midpoint of the chord has x = (x₁ + x₂)/2.
4
Chord length
|AB| = √(1 + k²) · √((x₁ + x₂)² − 4x₁x₂) — no need to find the roots themselves.

|AB| = √(1 + k²) · |x₁ − x₂|

where:

kslope of the line
x₁, x₂x-coordinates of A and B

The same chord formula works for any curve, the parabola and the hyperbola included; |x₁ − x₂| comes from Vieta’s formulas without solving the equation.

Example 10: a line and an ellipse

The line y = x + m and the ellipse x²/4 + y² = 1 are given.
1) For which m do they have two common points?
2) Find the length of the chord for m = 1.
3) Find the midpoint of that chord.

Show solution

Substitute: x²/4 + (x + m)² = 1, i.e. 5x² + 8mx + 4m² − 4 = 0.
1) Δ = 64m² − 20(4m² − 4) = 80 − 16m² > 0, so m² < 5: −√5 < m < √5 (m = ±√5 gives the two tangents).
2) For m = 1: 5x² + 8x = 0, x₁ = 0, x₂ = −8/5, and |AB| = √2 · 8/5 = 8√2/5.
3) x = (x₁ + x₂)/2 = −4/5 and y = x + 1 = 1/5: the midpoint is (−4/5, 1/5).

Example 11 (CSCA style): lines that miss the ellipse

The line y = kx + 2 and the ellipse x²/2 + y² = 1 have no common points. What is the range of k?
A) −√6/2 < k < √6/2 B) k < −√6/2 or k > √6/2 C) −√2 < k < √2 D) k ≠ 0

Show solution

Substitute: x² + 2(kx + 2)² = 2, i.e. (1 + 2k²)x² + 8kx + 6 = 0. No common points means Δ = 64k² − 24(1 + 2k²) = 16k² − 24 < 0, so k² < 3/2 and −√6/2 < k < √6/2, answer A. Shortcut: m² = a²k² + b² gives the tangents: 4 = 2k² + 1, k² = 3/2.
B is the secant case (Δ > 0). C swaps a² and b² in the tangent condition. A picture helps: (0, 2) lies outside the ellipse; flat lines through it miss the ellipse and steep ones cut it.

How CSCA asks about this

Standard equation: find a and c (or b) from the data; the foci are on the axis of the larger denominator; if their position is not given, write both orientations.
Parameters: x²/m + y²/n = 1 is an ellipse when m > 0, n > 0 and m ≠ n; its foci are on the x-axis when m > n.
Eccentricity: turn every condition into a relation between a, b and c, use a² = b² + c², and check that 0 < e < 1.
Definition and focal triangles: |PF₁| + |PF₂| = 2a; perimeters 2a + 2c and 4a; area b² tan(θ/2); the distance to a focus lies between a − c and a + c.
Line and ellipse: substitute, use Δ and Vieta’s formulas; chord √(1 + k²)|x₁ − x₂|; a tangent when m² = a²k² + b²; midpoint slope −b²x₀/(a²y₀).
Time-savers (about 75 s per item): test the options — a point given in the problem must satisfy the equation you choose; an e of 1 or more is never an ellipse; a quick sketch shows which axis is the major one.

The test can be taken in English or in Chinese. These are the terms of this lesson in both languages:

Term	中文	Pinyin
ellipse	椭圆	tuǒyuán
focus (plural foci)	焦点	jiāodiǎn
focal distance (2c)	焦距	jiāojù
major axis	长轴	chángzhóu
minor axis	短轴	duǎnzhóu
vertex (plural vertices)	顶点	dǐngdiǎn
eccentricity	离心率	líxīnlǜ
standard equation	标准方程	biāozhǔn fāngchéng
focal triangle	焦点三角形	jiāodiǎn sānjiǎoxíng
focal radius	焦半径	jiāobànjìng
locus (the path of a point)	轨迹	guǐjì
midpoint of a chord	弦的中点	xián de zhōngdiǎn
discriminant	判别式	pànbiéshì

Key points

Definition: |PF₁| + |PF₂| = 2a > |F₁F₂| = 2c; if 2a = 2c the path is the segment F₁F₂, if 2a < 2c there is no such point.
x²/a² + y²/b² = 1 or y²/a² + x²/b² = 1 with a > b > 0; the foci lie on the axis of the larger denominator.
a² = b² + c² (a is the hypotenuse); major axis 2a, minor axis 2b, focal distance 2c.
e = c/a = √(1 − b²/a²), 0 < e < 1; the larger e, the flatter the ellipse.
Focal triangle: perimeter 2a + 2c, over a focal chord 4a, area b² tan(θ/2); a − c ≤ |PF| ≤ a + c.
Line and ellipse: substitute and use Δ; chord √(1 + k²)|x₁ − x₂|; y = kx + m is a tangent when m² = a²k² + b².

Check yourself

12 questions. Every correct answer earns XP.

1 / 12

For the ellipse x²/a² + y²/b² = 1 with a > b > 0 and foci (±c, 0), which relation is true?

Topic test: 20 questions · 25 min

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