1. Chapter at a Glance
Slice a double cone with a flat plane and, depending on the angle of the cut, you get a circle, an ellipse, a parabola or a hyperbola. These conic sections, studied by the Greek geometer Apollonius (around 200 BCE), turn out to describe the real world: planets move in ellipses, a thrown ball follows a parabola, satellite dishes and car headlights use parabolic reflectors, and some comets follow hyperbolic paths. In this chapter we find the standard equations of each conic and their key features — focus, directrix, axes, latus rectum and eccentricity.
- Sections of a cone and degenerate conics.
- Circle: (x − h)² + (y − k)² = r².
- Parabola: focus–directrix definition; four standard forms; latus rectum.
- Ellipse: sum of focal distances; a, b, c; eccentricity.
- Hyperbola: difference of focal distances; a, b, c; eccentricity.
- Finding equations from given data and features from given equations.
2. Sections of a Cone
A double-napped cone has a vertical axis and a generator making a semi-vertical angle α. A plane cutting the cone at angle β with the axis gives:
| Angle of plane | Section |
|---|---|
| β = 90° (perpendicular to axis) | Circle |
| α < β < 90° | Ellipse |
| β = α (parallel to a generator) | Parabola |
| 0 ≤ β < α (cuts both nappes) | Hyperbola |
If the plane passes through the vertex, we get degenerate conics: a point, a line, or a pair of intersecting lines.
3. Circle
Centre (h, k), radius r: (x − h)² + (y − k)² = r². Centre at origin: x² + y² = r².
General form x² + y² + 2gx + 2fy + c = 0 has centre (−g, −f) and radius √(g² + f² − c).
Find the centre and radius of x² + y² + 8x + 10y − 8 = 0.
Solution: (x + 4)² + (y + 5)² = 8 + 16 + 25 = 49. Centre (−4, −5), radius 7.
Find the circle through (2, −2) and (3, 4) whose centre lies on x + y = 2.
Solution: Centre (h, 2 − h). Equal distances: (h − 2)² + (4 − h)² = (h − 3)² + (−2 − h)² → −12h + 20 = −2h + 13 → h = 0.7, k = 1.3. r² = (0.7 − 2)² + (1.3 + 2)² = 1.69 + 10.89 = 12.58. (x − 0.7)² + (y − 1.3)² = 12.58.
4. Parabola
A parabola is the set of points equidistant from a fixed point (the focus) and a fixed line (the directrix). Its eccentricity is e = 1.
| Equation | Opens | Focus | Directrix | Axis |
|---|---|---|---|---|
| y² = 4ax | Right | (a, 0) | x = −a | x-axis |
| y² = −4ax | Left | (−a, 0) | x = a | x-axis |
| x² = 4ay | Up | (0, a) | y = −a | y-axis |
| x² = −4ay | Down | (0, −a) | y = a | y-axis |
The latus rectum (chord through the focus perpendicular to the axis) has length 4a. Vertex at the origin.
Find the focus, directrix and latus rectum of y² = 8x.
Solution: 4a = 8 → a = 2. Focus (2, 0), directrix x = −2, latus rectum 8.
Find the parabola with vertex (0, 0), axis along the y-axis, passing through (2, −3).
Solution: x² = 4ay with (2, −3): 4 = −12a → a = −1/3 → x² = −(4/3)y, i.e. 3x² = −4y.
5. Ellipse
An ellipse is the set of points whose distances from two fixed points (the foci) have a constant sum 2a.
Major axis along x: x²/a² + y²/b² = 1, a > b, with c² = a² − b²; foci (±c, 0); vertices (±a, 0).
Eccentricity e = c/a, 0 < e < 1. Latus rectum = 2b²/a.
If the major axis is along y: x²/b² + y²/a² = 1 (a > b), foci (0, ±c).
Find the foci, vertices, lengths of axes, eccentricity and latus rectum of x²/25 + y²/9 = 1.
Solution: a = 5, b = 3, c = 4. Foci (±4, 0); vertices (±5, 0); major axis 10, minor axis 6; e = 4/5; latus rectum = 2 × 9/5 = 18/5.
Find the ellipse with foci (0, ±5) and major axis of length 26.
Solution: a = 13, c = 5 → b² = 169 − 25 = 144. Major axis along y: x²/144 + y²/169 = 1.
6. Hyperbola
A hyperbola is the set of points whose distances from two foci have a constant difference 2a (in absolute value).
Transverse axis along x: x²/a² − y²/b² = 1, with c² = a² + b²; foci (±c, 0); vertices (±a, 0).
Eccentricity e = c/a > 1. Latus rectum = 2b²/a.
If the transverse axis is along y: y²/a² − x²/b² = 1, foci (0, ±c). When a = b, the hyperbola is rectangular (e = √2).
Find the foci, vertices, eccentricity and latus rectum of 9y² − 4x² = 36.
Solution: y²/4 − x²/9 = 1: a = 2, b = 3, c = √13. Foci (0, ±√13); vertices (0, ±2); e = √13/2; latus rectum = 2 × 9/2 = 9.
7. Solved Problems (Board Pattern)
A parabolic reflector is 20 cm across and 5 cm deep. Find its focus.
Solution: Vertex at origin, axis along x: y² = 4ax through (5, 10): 100 = 20a → a = 5. Focus at 5 cm from the vertex on the axis.
Find the hyperbola with vertices (±7, 0) and e = 4/3.
Solution: a = 7, c = 28/3, b² = 784/9 − 49 = 343/9. x²/49 − 9y²/343 = 1.
Find the ellipse with major axis along the x-axis passing through (4, 3) and (6, 2).
Solution: 16/a² + 9/b² = 1, 36/a² + 4/b² = 1. Let p = 1/a², q = 1/b²: 16p + 9q = 1, 36p + 4q = 1 → p = 1/52, q = 1/13. x²/52 + y²/13 = 1.
8. Master Formula Table
| Conic | Standard equation | Key relation | e | Latus rectum |
|---|---|---|---|---|
| Circle | (x − h)² + (y − k)² = r² | — | 0 | — |
| Parabola | y² = 4ax | Focus (a, 0), directrix x = −a | 1 | 4a |
| Ellipse | x²/a² + y²/b² = 1 | c² = a² − b² | c/a < 1 | 2b²/a |
| Hyperbola | x²/a² − y²/b² = 1 | c² = a² + b² | c/a > 1 | 2b²/a |
9. Master Summary Table
| Conic | Defining property | Real-life example |
|---|---|---|
| Circle | Fixed distance from centre | Wheels |
| Parabola | PS = distance to directrix | Satellite dish, projectile path |
| Ellipse | PF₁ + PF₂ = 2a | Planetary orbits |
| Hyperbola | |PF₁ − PF₂| = 2a | Some comet paths, cooling-tower shape |
Chart — eccentricity of the conics:
10. Mind Map
- Centre, radius
- Complete squares
- Focus, directrix
- LR = 4a
- c² = a² − b²
- e < 1
- c² = a² + b²
- e > 1
11. Tips and Tricks
- For an ellipse, the larger denominator tells the major-axis direction.
- For a hyperbola, the positive term tells the transverse-axis direction.
- For a parabola, the squared variable tells the axis: y² → axis along x.
- Using c² = a² + b² for an ellipse (it is a² − b²).
- Taking a from the smaller denominator of an ellipse.
- Forgetting to divide to make the right-hand side 1 before reading a and b.
12. Important One-liners
- A plane parallel to a generator of the cone gives a parabola.
- Eccentricity: circle 0, ellipse < 1, parabola = 1, hyperbola > 1.
- The latus rectum of y² = 4ax is 4a.
- For an ellipse, sum of focal distances = major axis length 2a.
- A rectangular hyperbola has e = √2.
13. Exam-pattern Questions with Detailed Answers
The radius of x² + y² − 4x − 8y − 45 = 0 is:
- √65
- 5
- 9
- √45
Show answer
(x − 2)² + (y − 4)² = 45 + 4 + 16 = 65 → r = √65.
Answer: (A)
Find the focus and directrix of x² = −16y.
Show answer
4a = 16 → a = 4, opens down: focus (0, −4), directrix y = 4.
Answer: (0, −4); y = 4
Write the equation of the circle with centre (−2, 3) and radius 4.
Show answer
(x + 2)² + (y − 3)² = 16.
Answer: (x + 2)² + (y − 3)² = 16
Find the eccentricity and foci of x²/36 + y²/16 = 1.
Show answer
a = 6, b = 4, c = √20 = 2√5. e = √5/3; foci (±2√5, 0).
Answer: e = √5/3; (±2√5, 0)
Find the parabola with focus (6, 0) and directrix x = −6.
Show answer
a = 6: y² = 24x.
Answer: y² = 24x
Find the hyperbola with foci (0, ±12) and latus rectum 36.
Show answer
c = 12, 2b²/a = 36 → b² = 18a. a² + 18a = 144 → a = 6, b² = 108. y²/36 − x²/108 = 1.
Answer: y²/36 − x²/108 = 1
Find the ellipse with vertices (±5, 0) and foci (±4, 0).
Show answer
a = 5, c = 4, b² = 9: x²/25 + y²/9 = 1.
Answer: x²/25 + y²/9 = 1
Find the circle with centre (2, 2) passing through (4, 5).
Show answer
r² = 4 + 9 = 13: (x − 2)² + (y − 2)² = 13.
Answer: (x − 2)² + (y − 2)² = 13
A rod AB of length 12 cm moves with its ends always touching the coordinate axes. Find the locus of the point P on AB that is 3 cm from the end touching the x-axis.
Show answer
Let the rod make angle θ with the x-axis. P = (9 cos θ, 3 sin θ) (9 cm from the y-axis end horizontally, 3 cm vertically). cos θ = x/9, sin θ = y/3 → x²/81 + y²/9 = 1, an ellipse.
Answer: x²/81 + y²/9 = 1
An arch is a semi-ellipse 8 m wide at the base and 2 m high at the centre. Find the height of the arch 1.5 m from one end.
Show answer
x²/16 + y²/4 = 1. 1.5 m from an end means x = 2.5: y² = 4(1 − 6.25/16) = 4 × 9.75/16 = 2.4375 → y ≈ 1.56 m.
Answer: ≈ 1.56 m
Find the equation of the set of points P such that the sum of distances from (4, 0) and (−4, 0) is 10.
Show answer
Ellipse with 2a = 10, c = 4 → b² = 9: x²/25 + y²/9 = 1.
Answer: x²/25 + y²/9 = 1
Find the area of the triangle formed by the vertex of y² = 12x and the ends of its latus rectum.
Show answer
a = 3; latus rectum ends (3, ±6). Triangle with vertex (0, 0): base 12, height 3 → area 18 sq units.
Answer: 18 sq units
14. Quick Revision Summary
- Conics come from cutting a double cone; e = 0, < 1, = 1, > 1.
- Circle: (x − h)² + (y − k)² = r²; complete squares.
- Parabola y² = 4ax: focus (a, 0), directrix x = −a, LR 4a.
- Ellipse x²/a² + y²/b² = 1: c² = a² − b², e = c/a, LR 2b²/a.
- Hyperbola x²/a² − y²/b² = 1: c² = a² + b², e = c/a, LR 2b²/a.