Maths › Further Pure 1 › Conic sections
Conic sections
Parabola, ellipse and hyperbola are one family: the points whose distance from a focus is a fixed multiple of their distance from a line. That multiple, the eccentricity, decides which curve appears.
Builds on Parametric equations and Circles.
IN THIS TOPIC
- Quote the cartesian and parametric forms of the four standard conics.
- Compute eccentricity, foci and directrices from the curve's equation.
- Use the focus-directrix property to solve distance problems.
WHAT YOU PROBABLY THINK
The parabola, ellipse and hyperbola are three unrelated curves that happen to share a chapter.
The standard four
Each conic comes with a cartesian equation and a parametrisation worth knowing cold. The parabola y² = 4ax is (at², 2at); the ellipse:
the hyperbola x²/a² − y²/b² = 1 is (a sec t, b tan t) or (±a cosh t, b sinh t), and the rectangular hyperbola xy = c² is (ct, c/t). The parametric forms turn locus questions into single-variable algebra, which is why examiners lead with them.
WORKED EXAMPLE
Reading an ellipse
For the ellipse x²/25 + y²/9 = 1, find the eccentricity, foci and directrices.
a = 5, b = 3, and b² = a²(1 − e²) gives 9 = 25(1 − e²), so e² = 16/25 and e = 4/5.
Foci (±ae, 0) = (±4, 0); directrices x = ±a/e = ±25/4.
Check with the point (0, 3): distances to the two foci are 5 and 5, and their sum 10 equals 2a, as it must everywhere on the ellipse.
One property, three curves
The focus-directrix property defines the whole family: distance to the focus equals e times distance to the directrix. e < 1 closes the curve into an ellipse, e = 1 balances it into a parabola, e > 1 splits it into a hyperbola's two branches. Far from being unrelated, the three curves are one definition with a dial, and the opening claim mistakes the dial's settings for different machines.
WORKED EXAMPLE
A parabola's defining balance
For y² = 12x, state the focus and directrix, and verify the defining property at the point (3, 6).
4a = 12, so a = 3: focus (3, 0), directrix x = −3.
Distance from (3, 6) to the focus: 6. Distance to the directrix: 3 + 3 = 6.
Equal, as e = 1 demands: every point of a parabola sits exactly as far from the focus as from the directrix.
YOUR TURN
A hyperbola's constants
For x²/9 − y²/16 = 1, find e and the foci, and evaluate the difference of focal distances at the vertex (3, 0).
Show the working
b² = a²(e² − 1): 16 = 9(e² − 1), so e² = 25/9 and e = 5/3.
Foci (±ae, 0) = (±5, 0).
From (3, 0): distances 8 and 2, difference 6 = 2a. The constant difference of focal distances is the hyperbola's version of the ellipse's constant sum.
THE EXAM BIT
- Learn which formula carries the minus: b² = a²(1 − e²) for the ellipse, a²(e² − 1) for the hyperbola.
- Foci and directrices come as symmetric pairs; quoting only the positive one drops a mark.
- For the parabola, a is read from y² = 4ax; halving 4a instead of quartering it is the standard slip.
- Parametrise before chasing a locus; one parameter beats two coordinates.
CHECK YOURSELF
Write down parametric coordinates for a general point on y² = 8x and on xy = 9.
Show a hint
y² = 4ax has (at², 2at); xy = c² has (ct, c/t).
Show the answer
y
²
=
8
x
h
a
s
a
=
2
:
t
h
e
p
o
i
n
t
(
2
t
²
,
4
t
)
.
x
y
=
9
h
a
s
c
=
3
:
t
h
e
p
o
i
n
t
(
3
t
,
3
/
t
)
.
Parabola (at², 2at); ellipse (a cos t, b sin t); hyperbola (a sec t, b tan t); rectangular hyperbola (ct, c/t).
Distance to focus = e × distance to directrix: e < 1 ellipse, e = 1 parabola, e > 1 hyperbola.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
CHECK YOUR PROGRESS
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- Quote the cartesian and parametric forms of the four standard conics.
- Compute eccentricity, foci and directrices from the curve's equation.
- Use the focus-directrix property to solve distance problems.
Open the full revision checklist to see every objective in the course in one place.
No animated video for this topic yet; these notes stand alone.