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, determines which curve you get.
Builds on Parametric equations and Circles.
IN THIS TOPIC
- Find the cartesian and parametric forms of the four standard conics in the booklet and use them.
- Compute eccentricity, foci and directrices from a curve's equation.
- Use the focus-directrix property to solve distance problems.
- Write down the asymptotes of a hyperbola and use them when sketching.
- Handle an ellipse whose major axis runs vertically.
COMMON MISCONCEPTION
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, and the booklet prints all four of each in a single table. The parabola y² = 4ax is (at², 2at), and the ellipse pairs with a sine and a cosine.
The hyperbola x²/a² − y²/b² = 1 is (a sec t, b tan t), or (±a cosh t, b sinh t) with the sign choosing a branch. The rectangular hyperbola xy = c² is (ct, c/t). All of that is on the page in front of you in the exam, so spend your effort on using the forms rather than storing them. Parametric forms turn locus questions into single-variable algebra, and examiners lead with them for that reason.
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). Its distances to the two foci are 5 and 5, and the 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. Take e < 1 and the curve closes into an ellipse, e = 1 balances it into a parabola, and e > 1 splits it into a hyperbola's two branches. Three curves, one definition with a dial on it.
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. The focus is (3, 0) and the directrix x = −3.
Distance from (3, 6) to the focus is 6. Distance to the directrix is 3 + 3 = 6.
Equal, as e = 1 demands. Every point of a parabola sits exactly as far from the focus as from the directrix.
Asymptotes, orientation and sketching
A hyperbola runs away along two straight lines, and you can find them in one step. Replace the 1 in x²/a² − y²/b² = 1 by 0 and factorise, which gives y = ±(b/a)x. For large x the constant stops mattering, and the curve settles onto that pair. Sketch the asymptotes first, mark the vertices at (±a, 0), and the shape draws itself.
The rectangular hyperbola is the case a = b, where the asymptotes meet at right angles and the eccentricity is √2. Rotate it by 45° and it becomes the familiar xy = c², whose asymptotes are the coordinate axes.
Orientation catches people out. In x²/a² + y²/b² = 1 with b > a, the major axis is vertical, the foci sit at (0, ±be) on the y-axis, and the relation to use is a² = b²(1 − e²). Look at which denominator is larger before quoting anything, because writing e from the wrong one gives an eccentricity above 1 for a closed curve, which should stop you immediately.
GUIDED PRACTICE
A hyperbola's constants
For x²/9 − y²/16 = 1, find e, the foci and the asymptotes, 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), and the asymptotes are y = ±4x/3.
From (3, 0) the distances are 8 and 2, differing by 6 = 2a. That constant difference is the hyperbola's version of the ellipse's constant sum.
ASSESSMENT FOCUS
- Do not memorise which relation carries the minus sign. The booklet's conics table has a whole eccentricity row, with b² = a²(1 − e²) under the ellipse and b² = a²(e² − 1) under the hyperbola. Read across.
- Foci and directrices come in symmetric pairs, and 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.
- Check which denominator is larger on an ellipse before deciding where the foci lie.
- Draw the asymptotes before the branches when sketching a hyperbola.
- Parametrise before working out a locus, since one parameter is easier to handle than 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
y² = 8x has a = 2, giving the point (2t², 4t). xy = 9 has c = 3, giving the point (3t, 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, with e < 1 an ellipse, e = 1 a parabola, e > 1 a hyperbola.
Foci sit at (±ae, 0) and directrices at x = ±a/e when the major axis is horizontal.
The hyperbola x²/a² − y²/b² = 1 has asymptotes y = ±(b/a)x.
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.
Or read them with their worked answers on the conic sections questions page.
CHECK YOUR PROGRESS
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- Find the cartesian and parametric forms of the four standard conics in the booklet and use them.
- Compute eccentricity, foci and directrices from a curve's equation.
- Use the focus-directrix property to solve distance problems.
- Write down the asymptotes of a hyperbola and use them when sketching.
- Handle an ellipse whose major axis runs vertically.
Open the full revision checklist to see every objective in the course in one place.