MathematicsCore22 min read

Conic Sections

Four curves, one cone, and the number that tells them apart

This topic appears in:

2nd Year Mathematics · ch.6
01

One cone, four curves

Definition

Eccentricity — For a fixed point (the focus) and a fixed line (the directrix), a conic is the set of points whose distance from the focus divided by its distance from the directrix equals a constant e. That constant is the eccentricity: e = 0 gives a circle, e < 1 an ellipse, e = 1 a parabola and e > 1 a hyperbola.

Take a double cone and slice it with a plane. The shape of the cut depends entirely on the angle of the slice. Cut horizontally and you get a circle. Tilt the plane a little and the circle stretches into an ellipse. Tilt it until it runs parallel to the side of the cone and the curve opens out and never closes — a parabola. Tilt further, so the plane cuts both halves of the cone, and you get a hyperbola in two branches.

These four curves are the conic sections, and they are not four unrelated topics. One definition covers all of them, and one number decides which you have.

Move through the four options and watch the eccentricity in the read-out. On the ellipse, push b towards a and e falls towards 0 — the two foci merge and the ellipse becomes a circle.

02

The circle

The simplest conic: every point the same distance from the centre. Its equation comes straight from the distance formula, and the general form is what you get after expanding.

Given a general equation, complete the square in x and in y to recover the centre and the radius. If the resulting right-hand side is negative there is no circle at all, and saying so is the answer.

(x − h)² + (y − k)² = r²centre (h, k), radius rx² + y² + 2gx + 2fy + c = 0centre (−g, −f), r = √(g² + f² − c)tangent at (x₁, y₁) on x² + y² = r²: xx₁ + yy₁ = r²a general second-degree equation is a circle only when the x² and y² coefficients are equal and there is no xy term
Worked example

Find the centre and radius of x² + y² − 6x + 4y − 12 = 0.

  1. Group and complete the square in x: x² − 6x = (x − 3)² − 9.Halve the coefficient of x, square it, add and subtract.
  2. And in y: y² + 4y = (y + 2)² − 4.
  3. Substituting: (x − 3)² − 9 + (y + 2)² − 4 − 12 = 0, so (x − 3)² + (y + 2)² = 25.Collect the three constants on the right.
  4. Centre (3, −2), radius 5.Or use the shortcut: 2g = −6 so g = −3, 2f = 4 so f = 2, giving centre (3, −2) and r = √(9 + 4 + 12) = 5 ✓

Centre (3, −2), radius 5

03

The parabola

A parabola is the set of points equidistant from the focus and the directrix — so e = 1 exactly. That single property explains its most useful behaviour: every ray arriving parallel to the axis is reflected through the focus, which is why satellite dishes and reflecting telescopes are parabolic.

The latus rectum is the chord through the focus perpendicular to the axis; its length, 4a, tells you how wide the curve opens.

y² = 4axfocus (a, 0), directrix x = −a, opens righty² = −4axopens leftx² = 4ayfocus (0, a), directrix y = −a, opens uplatus rectum = 4athe variable that is NOT squared tells you the axis the parabola opens along

Reading a parabola in one glance

Whichever variable is squared, the curve is symmetric about the other axis. In y² = 4ax the y is squared, so it is symmetric about the x-axis and opens left or right; the sign of the term decides which. In x² = 4ay it opens up or down. Getting this right takes two seconds and prevents a completely wrong sketch.

04

The ellipse and the hyperbola

Both have two foci, and both are defined by a condition on the distances to them. For an ellipse the sum of the two distances is constant — which is why the gardener's method works: pin a loop of string at two points and trace it taut. For a hyperbola the difference is constant.

The equations differ by one sign, and that sign changes everything: a plus gives a closed oval, a minus gives two branches running off to infinity along asymptotes.

ellipse: x²/a² + y²/b² = 1, c² = a² − b², e = c/a < 1hyperbola: x²/a² − y²/b² = 1, c² = a² + b², e = c/a > 1asymptotes of the hyperbola: y = ±(b/a) xfoci at (±c, 0) in both casesnote the sign inside c²: subtract for the ellipse, add for the hyperbola
a
semi-major axis (ellipse) or the distance to a vertexalways the larger for an ellipse
b
semi-minor axis (ellipse) or the conjugate semi-axis
c
distance from the centre to a focus
Worked example

For the ellipse x²/25 + y²/9 = 1, find the vertices, the foci and the eccentricity.

  1. a² = 25 and b² = 9, so a = 5 and b = 3. Since a > b the major axis lies along the x-axis.The larger denominator sits under the variable along the major axis — that is how you tell which way the ellipse is oriented.
  2. Vertices at (±5, 0) and the ends of the minor axis at (0, ±3).Vertices are the ends of the major axis only.
  3. c² = a² − b² = 25 − 9 = 16, so c = 4 and the foci are at (±4, 0).Subtract for an ellipse. Adding would give the hyperbola relationship and put the foci outside the curve.
  4. e = c/a = 4/5 = 0.8.Less than 1, as it must be for an ellipse. A value of 0.8 means a distinctly stretched oval.

Vertices (±5, 0); foci (±4, 0); e = 0.8

The foci are always on the major axis

If b² > a² in the form given, the major axis is vertical, the roles of a and b swap, and the foci sit at (0, ±c). Check which denominator is larger before writing down a single coordinate. Assuming the foci are always on the x-axis is the most reliable way to lose every mark after the first.

Before you leave this chapter

  1. e = 0 circle, e < 1 ellipse, e = 1 parabola, e > 1 hyperbola.
  2. Circle: complete the square to find the centre and radius; a negative right-hand side means no circle.
  3. Parabola: the variable NOT squared gives the axis it opens along; latus rectum = 4a.
  4. Ellipse c² = a² − b²; hyperbola c² = a² + b². The sign is the whole difference.
  5. The foci lie on the major axis — check which denominator is larger before locating them.

Practice questions

6 questions · 20 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

3 · 6 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
Define the eccentricity of a conic and state its value for each of the four conics.
Model answer

Eccentricity is the constant ratio of a point's distance from the focus to its distance from the directrix. e = 0 circle, 0 < e < 1 ellipse, e = 1 parabola, e > 1 hyperbola.

Examiner tip. One definition and four values — the mark scheme usually splits exactly that way.

SQ2[2 marks]
Find the centre and radius of x² + y² + 8x − 6y = 0.
Model answer

Here 2g = 8 and 2f = −6, so g = 4, f = −3, c = 0. Centre (−g, −f) = (−4, 3) and radius √(16 + 9 − 0) = 5.

Examiner tip. A constant term of zero means the circle passes through the origin — a useful check, since (0,0) does satisfy the equation.

SQ3[2 marks]
State the focus and directrix of the parabola y² = 12x.
Model answer

Comparing with y² = 4ax gives 4a = 12, so a = 3. The focus is at (3, 0) and the directrix is the line x = −3.

Examiner tip. Divide by 4 to get a. Quoting a = 12 is the standard error and puts the focus four times too far out.

Solved numericals

2 · 8 marks

Full working, one step per line, with the marks shown where they are awarded.

N1[4 marks]
Find the foci and eccentricity of the hyperbola x²/16 − y²/9 = 1.
Full working
  1. a² = 16 and b² = 9, so a = 4, b = 3[1]
  2. For a hyperbola c² = a² + b² = 16 + 9 = 25add, unlike the ellipse[1]
  3. c = 5, so the foci are at (±5, 0)[1]
  4. e = c/a = 5/4 = 1.25greater than 1, as required for a hyperbola[1]

Foci (±5, 0); e = 1.25

Examiner tip. Check the eccentricity against the curve type before moving on. An ellipse answer above 1, or a hyperbola answer below it, means the sign in c² went the wrong way.

N2[4 marks]
Find the equation of the circle with centre (2, −3) that passes through the point (5, 1).
Full working
  1. The radius is the distance from the centre to the given point[1]
  2. r = √[(5 − 2)² + (1 + 3)²] = √(9 + 16) = 5[1]
  3. Standard form (x − 2)² + (y + 3)² = 25the sign of k flips inside the bracket[1]
  4. Expanding: x² + y² − 4x + 6y − 12 = 0accept either form unless one is specified[1]

(x − 2)² + (y + 3)² = 25, i.e. x² + y² − 4x + 6y − 12 = 0

Examiner tip. A centre of (2, −3) gives brackets (x − 2) and (y + 3). The sign inside the bracket is always the opposite of the coordinate.

Long questions

1 · 6 marks

Theory and numerical together, as they appear in the long-question section.

LQ1[6 marks]
Consider the ellipse 9x² + 25y² = 225.
  1. Write the equation in standard form and state a and b.
  2. Find the coordinates of the vertices and the foci.
  3. Find the eccentricity and the length of the latus rectum, given that it equals 2b²/a.
Mark scheme
  1. Divide throughout by 225: x²/25 + y²/9 = 1the right-hand side must be 1 before anything can be read off[1]
  2. a = 5, b = 3, and since a > b the major axis is along the x-axis[1]
  3. Vertices (±5, 0)the ends of the major axis[1]
  4. c² = 25 − 9 = 16, so foci at (±4, 0)subtract for an ellipse[1]
  5. e = 4/5 = 0.8less than 1 ✓[1]
  6. Latus rectum = 2(9)/5 = 3.6[1]

(a) x²/25 + y²/9 = 1, a = 5, b = 3 (b) vertices (±5,0), foci (±4,0) (c) e = 0.8, latus rectum 3.6

Examiner tip. Always divide through to make the right-hand side 1 first. Reading a² = 9 straight off 9x² + 25y² = 225 is the single most expensive mistake in this chapter.