Sections of a cone
A double cone is made by turning a slanted line (the generator) around a fixed axis. The two halves are called nappes. Cut it with a plane that does not pass through the tip (vertex):
- Plane at 90° to the axis → circle.
- Plane tilted a little → ellipse (a closed oval).
- Plane parallel to a generator → parabola (one open curve).
- Plane tilted even more, cutting both nappes → hyperbola (two open branches).
Degenerate conic sections
If the plane passes through the vertex, the cut is "squashed" into a simpler shape:
- like the circle/ellipse cut → just a point;
- like the parabola cut → a single straight line (a generator);
- like the hyperbola cut → a pair of intersecting lines.
Circle
A circle is the set of all points at a fixed distance r (radius) from a fixed point C(h, k) (centre). By the distance formula: (x − h)² + (y − k)² = r². Centre at the origin: x² + y² = r².
General form: x² + y² + 2gx + 2fy + c = 0, with centre (−g, −f) and radius √(g² + f² − c). Complete the squares to find them.
Parabola
A parabola is the set of points equally far from a fixed point (focus) and a fixed line (directrix). The line through the focus perpendicular to the directrix is the axis; the point halfway is the vertex.
| Equation | Focus | Directrix | Opens |
|---|---|---|---|
| y² = 4ax | (a, 0) | x = −a | right |
| y² = −4ax | (−a, 0) | x = a | left |
| x² = 4ay | (0, a) | y = −a | up |
| x² = −4ay | (0, −a) | y = a | down |
The latus rectum is the chord through the focus perpendicular to the axis. Its length is 4a.
Ellipse
An ellipse is the set of points whose distances from two fixed points (foci) add up to a constant 2a. With centre at the origin and foci on the x-axis: x²/a² + y²/b² = 1, a > b, and c² = a² − b².
- Foci (±c, 0); vertices (±a, 0); major axis 2a, minor axis 2b.
- Eccentricity e = c/a, and 0 < e < 1. Nearly 0 → almost a circle.
- Latus rectum = 2b²/a.
- If the foci are on the y-axis: x²/b² + y²/a² = 1 (the bigger number is under y²).
Hyperbola
A hyperbola is the set of points whose distances from two foci differ by a constant 2a. Standard form: x²/a² − y²/b² = 1, with c² = a² + b².
- Foci (±c, 0); vertices (±a, 0); transverse axis 2a, conjugate axis 2b.
- Eccentricity e = c/a > 1.
- Latus rectum = 2b²/a.
- Foci on the y-axis: y²/a² − x²/b² = 1 (the positive term tells the axis).
- The branches get close to the lines y = ±(b/a)x.
Key formulas and definitions
- (x − h)² + (y − k)² = r²
- x² + y² + 2gx + 2fy + c = 0: centre (−g, −f), r = √(g² + f² − c)
- y² = 4ax: focus (a, 0), directrix x = −a, latus rectum 4a
- x²/a² + y²/b² = 1: c² = a² − b², e = c/a < 1
- x²/a² − y²/b² = 1: c² = a² + b², e = c/a > 1
- Latus rectum of ellipse / hyperbola = 2b²/a
Worked examples
1. Find the equation of the circle with centre (−2, 3) and radius 4.
Step 1: (x + 2)² + (y − 3)² = 16. Step 2: x² + 4x + 4 + y² − 6y + 9 = 16. Step 3: x² + y² + 4x − 6y − 3 = 0.
2. Find the centre and radius of x² + y² + 8x − 10y − 8 = 0.
Step 1: (x² + 8x + 16) + (y² − 10y + 25) = 8 + 16 + 25. Step 2: (x + 4)² + (y − 5)² = 49. Step 3: centre (−4, 5), radius 7.
3. Find the focus, directrix and latus rectum of y² = 12x.
Step 1: 4a = 12, so a = 3. Step 2: focus (3, 0), directrix x = −3. Step 3: latus rectum = 4a = 12.
4. Find the focus and directrix of x² = −16y.
Step 1: form x² = −4ay with 4a = 16, a = 4. Step 2: opens downward: focus (0, −4). Step 3: directrix y = 4.
5. Find the parabola with vertex at the origin, axis along the x-axis, passing through (2, 3).
Step 1: form y² = 4ax (point has x > 0). Step 2: 9 = 4a × 2 → 4a = 9/2. Step 3: y² = 9x/2, or 2y² = 9x.
6. For x²/25 + y²/9 = 1, find foci, eccentricity and latus rectum.
Step 1: a = 5, b = 3. Step 2: c = √(25 − 9) = 4 → foci (±4, 0). Step 3: e = 4/5; latus rectum = 2 × 9/5 = 18/5.
7. Find the ellipse with vertices (±13, 0) and foci (±5, 0).
Step 1: a = 13, c = 5. Step 2: b² = a² − c² = 169 − 25 = 144. Step 3: x²/169 + y²/144 = 1.
8. For 9y² − 4x² = 36, find the foci and eccentricity.
Step 1: divide by 36: y²/4 − x²/9 = 1, axis along y. Step 2: a = 2, b = 3, c = √(4 + 9) = √13. Step 3: foci (0, ±√13), e = √13/2.
9. Find the hyperbola with foci (±5, 0) and transverse axis 8.
Step 1: 2a = 8 → a = 4; c = 5. Step 2: b² = c² − a² = 25 − 16 = 9. Step 3: x²/16 − y²/9 = 1.
Common mistakes
- Using c² = a² + b² for an ellipse. For an ellipse it is a² − b²; plus is only for the hyperbola.
- Reading the axis wrongly: in an ellipse the larger denominator decides the major axis; in a hyperbola the positive term decides it.
- Taking a = 12 in y² = 12x. Here 4a = 12, so a = 3.
- Forgetting to move the constant when completing squares, so the radius comes out wrong.