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Conic sections: Class 12 Mathematics
The circle, parabola, ellipse and hyperbola as sections of a cone: definitions, standard equations, centre, foci, vertices, directrix, eccentricity, latus rectum, and finding equations from given data.
Key points
A conic section is the curve formed when a plane cuts a right circular cone. Depending on the angle of the plane, the curve is a circle, a parabola, an ellipse or a hyperbola. Each conic can also be described as the locus of a point P whose distance from a fixed point (the focus) is e times its distance from a fixed line (the directrix). The constant e is the eccentricity: e = 0 for a circle, e < 1 for an ellipse, e = 1 for a parabola, e > 1 for a hyperbola.
Circle: the set of points at a fixed distance r (the radius) from a fixed point C(h, k), the centre. Its standard equation is (x − h)² + (y − k)² = r². With centre at the origin it is x² + y² = r².
General equation of a circle: x² + y² + 2gx + 2fy + c = 0. The centre is (−g, −f) and the radius is r = √(g² + f² − c). The circle is real when g² + f² − c > 0. A line meets a circle in two points, one point (a tangent) or no point, depending on the distance from the centre to the line compared with r.
Parabola: the locus of a point equidistant from a fixed point (focus) and a fixed line (directrix). The standard form y² = 4ax has focus (a, 0), directrix x = −a, vertex (0, 0) and axis the x-axis. It opens right for a > 0 and left for a < 0. The form x² = 4ay opens upward for a > 0, with focus (0, a) and directrix y = −a.
The latus rectum is the chord through the focus perpendicular to the axis. For the parabola y² = 4ax its length is 4a, and its end points are (a, 2a) and (a, −2a). The vertex form is (y − k)² = 4a(x − h), with vertex (h, k).
Ellipse: the locus of a point whose distances from two fixed points (foci) have a constant sum 2a. The standard equation with centre at the origin and the major axis along the x-axis is x²/a² + y²/b² = 1 with a > b > 0. Here c² = a² − b², the foci are (±c, 0), the vertices are (±a, 0), and the eccentricity is e = c/a, which is less than 1.
Features of the ellipse x²/a² + y²/b² = 1 (a > b):
- Major axis length 2a along the x-axis; minor axis length 2b along the y-axis.
- Directrices x = ±a/e.
- Length of the latus rectum is 2b²/a.
- If the larger denominator is under y², the major axis lies along the y-axis and the foci are (0, ±c).
Hyperbola: the locus of a point whose distances from two fixed points (foci) have a constant difference 2a. The standard equation is x²/a² − y²/b² = 1. Here c² = a² + b², the foci are (±c, 0), the vertices are (±a, 0), and e = c/a is greater than 1. The curve has two branches.
The asymptotes of x²/a² − y²/b² = 1 are the lines y = ±(b/a)x. The branches come closer and closer to them but never meet them. The length of the latus rectum is 2b²/a. The hyperbola y²/a² − x²/b² = 1 opens up and down, with foci (0, ±c).
To get the type of conic from a general second degree equation Ax² + Bxy + Cy² + Dx + Ey + F = 0 with B = 0: it is a circle if A = C, a parabola if only one of A or C is zero, an ellipse if A and C have the same sign and are unequal, and a hyperbola if A and C have opposite signs.
To find an equation, identify the type, write the standard form, and use the given data (focus, vertex, eccentricity, or points on the curve) to find the unknown constants a, b, h, k. To sketch, mark the centre or vertex, the foci, the axes, and a few points.
Study tip: Memorise a table with one row for each conic: equation, focus, vertex, directrix, eccentricity and latus rectum. Always compare with the standard form before reading off any value. Sketch the curve with the foci marked, as many questions award a mark for the diagram.
Definitions and flashcards
Say the meaning first, then open the card to check.
Locus
The path of a point that moves under a given condition.
Focus
A fixed point used to define a conic.
Directrix
A fixed line used to define a conic.
Eccentricity
The constant ratio e of the distance from the focus to the distance from the directrix.
Latus rectum
The chord through a focus perpendicular to the axis of the conic.
Vertex
A point where the conic meets its axis of symmetry.
Major axis
The longer axis of an ellipse, of length 2a.
Asymptote
A line that a curve approaches without meeting it.
Formula sheet
| Name | Formula | What it means |
|---|---|---|
| Circle | (x − h)² + (y − k)² = r² | Centre (h, k), radius r. |
| General circle | x² + y² + 2gx + 2fy + c = 0 | Centre (−g, −f); r = √(g² + f² − c). |
| Parabola | y² = 4ax | Focus (a, 0); directrix x = −a; latus rectum 4a. |
| Ellipse | x²/a² + y²/b² = 1 | c² = a² − b²; e = c/a < 1. |
| Hyperbola | x²/a² − y²/b² = 1 | c² = a² + b²; e = c/a > 1. |
| Latus rectum | 2b²/a | For the ellipse and the hyperbola. |
| Asymptotes | y = ±(b/a)x | For x²/a² − y²/b² = 1. |
Short questions with model answers
These are the kind of questions asked for 2, 3 and 5 marks. Try each one before opening the answer.
Define a parabola and write the standard equation with the focus on the x-axis. (2 marks)
A parabola is the locus of a point that is always at equal distance from a fixed point (the focus) and a fixed line (the directrix). Its standard equation is y² = 4ax, with focus (a, 0) and directrix x = −a.
State the value of eccentricity for each conic. (2 marks)
Circle: e = 0. Ellipse: 0 < e < 1. Parabola: e = 1. Hyperbola: e > 1.
Find the centre and radius of x² + y² − 6x + 4y − 12 = 0. (3 marks)
Compare with x² + y² + 2gx + 2fy + c = 0. Then 2g = −6, so g = −3, and 2f = 4, so f = 2, and c = −12. The centre is (−g, −f) = (3, −2). The radius is √(9 + 4 + 12) = √25 = 5.
Write the relation between a, b and c for an ellipse and for a hyperbola. (2 marks)
For the ellipse c² = a² − b². For the hyperbola c² = a² + b². Here c is the distance from the centre to a focus.
Explain how an ellipse differs from a hyperbola in definition and shape. (5 marks)
An ellipse is the locus of a point for which the sum of the distances from two fixed foci is constant (2a). A hyperbola is the locus of a point for which the difference of the distances from two foci is constant (2a). The ellipse is a closed oval curve with e < 1, and it has the equation x²/a² + y²/b² = 1 with c² = a² − b². The hyperbola has two open branches and e > 1, and it has the equation x²/a² − y²/b² = 1 with c² = a² + b². The hyperbola also has two asymptotes, y = ±(b/a)x, but the ellipse has none.
Numericals with worked steps
Find the equation of the circle with centre (2, −3) and radius 5.
- Standard form: (x − h)² + (y − k)² = r² with h = 2, k = −3, r = 5.
- (x − 2)² + (y + 3)² = 25.
- Expand: x² − 4x + 4 + y² + 6y + 9 = 25.
- x² + y² − 4x + 6y + 13 − 25 = 0, so x² + y² − 4x + 6y − 12 = 0.
Answer: (x − 2)² + (y + 3)² = 25, or x² + y² − 4x + 6y − 12 = 0
Find the centre and radius of the circle x² + y² − 6x + 4y − 12 = 0 by completing the square.
- Group: (x² − 6x) + (y² + 4y) = 12.
- Complete squares: (x² − 6x + 9) + (y² + 4y + 4) = 12 + 9 + 4.
- (x − 3)² + (y + 2)² = 25.
- Centre (3, −2), radius √25 = 5.
Answer: Centre (3, −2), radius 5 units
For the parabola y² = 12x find the focus, the directrix and the length of the latus rectum.
- Compare with y² = 4ax: 4a = 12, so a = 3.
- Focus (a, 0) = (3, 0).
- Directrix x = −a, so x = −3.
- Latus rectum length = 4a = 12.
Answer: Focus (3, 0); directrix x = −3; latus rectum 12 units
For the ellipse x²/25 + y²/9 = 1 find the foci, the eccentricity and the length of the latus rectum.
- a² = 25 so a = 5, and b² = 9 so b = 3.
- c² = a² − b² = 25 − 9 = 16, so c = 4.
- Foci (±4, 0).
- e = c/a = 4/5.
- Latus rectum = 2b²/a = 2(9)/5 = 18/5 = 3.6.
Answer: Foci (±4, 0); e = 4/5; latus rectum 18/5 units
For the hyperbola x²/16 − y²/9 = 1 find the foci, the eccentricity and the asymptotes.
- a² = 16 so a = 4, and b² = 9 so b = 3.
- c² = a² + b² = 16 + 9 = 25, so c = 5.
- Foci (±5, 0).
- e = c/a = 5/4.
- Asymptotes y = ±(b/a)x = ±(3/4)x.
Answer: Foci (±5, 0); e = 5/4; asymptotes y = ±3x/4
Common mistakes
- Using c² = a² − b² for the hyperbola, or c² = a² + b² for the ellipse. The signs are different.
- Writing the radius as g² + f² − c without the square root.
- Reading the focus of y² = 4ax as (4a, 0) instead of (a, 0).
- Taking a to be the larger denominator in the hyperbola. In a hyperbola a² is always under the positive term.
- Forgetting to change the signs of h and k when reading the centre from (x − h)² + (y − k)² = r².