Controls
Jump to a curve
Readings
- Curve
- Ellipse
- Tilt θ
- 30°
- Eccentricity e
- 0.707
- Rule
- 0 < e < 1
- Standard equation
- x²/3.00 + y²/1.50 = 1
- Semi-major axis a
- 1.732
- Semi-minor axis b
- 1.225
- Centre to focus
- 1.225
How to use this simulation
- Look at the left half: a double cone in 3D, with a translucent plane cutting through it and a bright orange curve where they meet. Drag the left half to spin the cone, or press play and it turns by itself. The right half draws that same orange curve flat.
- Use “Jump to a curve” for a quick start. Circle sets the tilt to 0°, ellipse to 30°, parabola to 45° and hyperbola to 65°. At the start the tilt is 30°, which cuts an ellipse.
- Drag the “Tilt of the plane” slider slowly from 0° to 80°. Watch the circle stretch into an ellipse, stretch further, and snap into a parabola at exactly 45°, the angle at which the plane is parallel to the side of the cone. Past 45° the single curve breaks into two branches: a hyperbola.
- Read the eccentricity e under the flat curve and in the readings. At 30° it is 0.707. It grows with the tilt: 0 for the circle, 1 at the parabola, bigger than 1 for the hyperbola.
- Move “Height where the plane crosses the axis”. The plane slides up or down the axis. The curve gets bigger or smaller, but its shape (and e) does not change, because the shape depends only on the tilt.
- Check the “Standard equation” reading. For the start position it shows x²/3.00 + y²/1.50 = 1, and the foci F and F′ sit 1.225 units from the centre.
A torch on a wall
Switch on a torch in a dark room and point it straight at a wall. The lit patch is a circle. Now tilt the torch upwards. The circle stretches into an oval. Tilt it more, and the oval runs off towards the ceiling, becoming an open, U-shaped edge. Tilt it past that and the lit region turns into a curve with two separate branches, one on the wall and (if you imagine the beam continuing backwards through the torch) one behind you.
That light beam is a cone, and the wall is a plane cutting it. The patch’s edge is a conic section. Greek mathematicians such as Menaechmus and Apollonius studied these curves more than two thousand years ago, long before anyone knew that planets move on them. This page lets you do the torch experiment with a slider.
Slicing the cone: one tilt, four curves
In the simulation the cone has a half-angle of 45°, so its side makes 45° with the flat base direction. The plane is tilted by an angle θ from the horizontal. The rule is short.
e = √2 · sin θeccentricity of the section for this 45° cone; 0 at θ = 0, exactly 1 at θ = 45°
- θ = 0°, plane perpendicular to the axis: circle.
- 0° < θ < 45°, flatter than the side of the cone: ellipse.
- θ = 45°, parallel to the side of the cone: parabola (one open branch).
- θ > 45°, steeper than the side: hyperbola (two branches, one in each half of the cone).
| Tilt θ | Curve | Eccentricity e |
|---|---|---|
| 0° | Circle | 0.000 |
| 15° | Ellipse | 0.366 |
| 30° | Ellipse | 0.707 |
| 45° | Parabola | 1.000 |
| 60° | Hyperbola | 1.225 |
| 75° | Hyperbola | 1.366 |
Eccentricity: how stretched is the curve?
A conic can also be defined without any cone. Pick a point F (the focus) and a straight line d (the directrix). A point P is on the conic when its distance to F is a fixed multiple e of its distance to d: PF = e × Pd. That multiple is the eccentricity.
Eccentricity is a stretchiness score. A circle has e = 0 and is perfectly round. As e rises towards 1 the ellipse gets longer and thinner. At exactly 1 it has been pulled open into a parabola. Above 1 it has split into the two arms of a hyperbola, and the larger e is, the flatter those arms.
PF = e × Pdthe focus-directrix definition
e = c / afor an ellipse (c = distance from centre to a focus, a = semi-major axis)
| Eccentricity | Curve |
|---|---|
| e = 0 | Circle |
| 0 < e < 1 | Ellipse |
| e = 1 | Parabola |
| e > 1 | Hyperbola |
The circle: the simplest conic
Every point on a circle is the same distance r from the centre. With the centre at the origin the standard equation is x² + y² = r². The two foci have merged into the centre, which is why its eccentricity is zero.
In the simulation, set the tilt to 0° and the circle’s radius equals the height c at which the plane crosses the axis. That is natural: the cone x² + y² = z² has radius equal to its height at every level.
x² + y² = r²centre at the origin, radius r
The ellipse: a stretched circle with two foci
An ellipse is the set of points whose distances to two fixed points (the foci F and F′) add up to a constant, 2a. The long half-width a is the semi-major axis and the short half-width b is the semi-minor axis. The foci lie on the long axis at a distance c from the centre, where c² = a² − b². The eccentricity is e = c ÷ a.
At the start of the simulation (tilt 30°, c = 1.0) the section is an ellipse with a² = 3.00 and b² = 1.50, so c² = 3.00 − 1.50 and the foci sit 1.225 from the centre; e = 1.225 ÷ 1.732 = 0.707, which matches √2 · sin 30°.
x²/a² + y²/b² = 1centre at the origin, a > b
c² = a² − b²foci at (±c, 0)
PF + PF′ = 2athe string definition
The parabola: the curve that never closes
A parabola is the set of points equally far from a focus F and a straight line, the directrix. With the vertex at the origin and the curve opening to the right, F = (a, 0), the directrix is x = −a, and the equation is y² = 4ax. Here a (often called p) is the focal length. The simulation draws the directrix as a dashed red line and reports p in the readings.
A parabola has no second focus and no centre; the reading “None” for a and b is honest. Its eccentricity is exactly 1, and that is the reason one tilt (45° here) is special: it is the borderline between curves that close up and curves that do not.
y² = 4axvertex at the origin, opening right, focus (a, 0), directrix x = −a
x² = 4ayopening upward, focus (0, a)
The hyperbola: two branches and two asymptotes
A hyperbola is the set of points where the difference of the distances to two foci is constant, 2a. It has two separate branches. Far from the centre each branch straightens towards a pair of lines called asymptotes, y = ±(b/a)x. The foci are farther from the centre than the vertices: c² = a² + b², so e = c ÷ a is always bigger than 1.
Try the preset. At 65° (c = 1) the readings show a = 0.657, b = 0.527 and e = 1.282. The 3D view shows why there are two branches: one lies in the upper half of the cone and the other in the lower half.
x²/a² − y²/b² = 1centre at the origin, opening left and right
c² = a² + b²foci at (±c, 0)
y = ±(b/a)xasymptotes
Name the curve from its equation
Exams often hand you an equation and ask which conic it is. Read the squares first. If x and y are both squared with the same sign and the same coefficient, think circle. Same sign, different coefficients: ellipse. Opposite signs: hyperbola. Only one of them squared: parabola.
For the general equation Ax² + Bxy + Cy² + Dx + Ey + F = 0, the discriminant B² − 4AC does it in one step: negative means an ellipse (a circle if A = C and B = 0), zero means a parabola, and positive means a hyperbola.
| Equation | What you see | Curve |
|---|---|---|
| x² + y² = 25 | same sign, equal coefficients | Circle |
| 4x² + 9y² = 36 | same sign, different coefficients | Ellipse |
| y² = 8x | only y is squared | Parabola |
| x² − y² = 4 | opposite signs | Hyperbola |
Try it in the simulation
Five quick experiments. Each one takes under a minute.
- Set the tilt to 0° and slowly raise it to 44°. Watch e climb from 0 and the ellipse stretch. Try 44° and 45° back to back: the ellipse reaches so far that it nearly escapes the cone.
- Press the Parabola button. The directrix appears. Move the plane height c: the parabola gets wider and p grows, but the shape stays a parabola.
- Press the Hyperbola button and drag the cone around. Find the two branches, one in each half of the cone.
- Raise the tilt from 46° to 80°. The two arms of the hyperbola close in towards a pair of crossing lines. At 90° the plane would contain the axis and cut the cone in two straight lines.
- Pause the spin, then compare the readings of a, b and the equation as you change the height c. Which of them change, and which stay the same?
Solved problems
Seven classic questions. Every number below is computed from the same formulas the simulation uses.
Problem 1: which curve does each tilt cut? (20°, 45°, 70°)
The cone’s side is at 45°, so compare each tilt with 45°. 20° is below it: ellipse, e = √2 · sin 20° = 0.484. 45° is equal to it: parabola, e = 1.000. 70° is above it: hyperbola, e = 1.329.
Problem 2: foci and eccentricity of x²/25 + y²/9 = 1
Compare with x²/a² + y²/b² = 1: a² = 25 so a = 5, and b² = 9 so b = 3. Then c² = a² − b² = 25 − 9 = 16, so c = 4.
The foci are (±4, 0). The eccentricity is e = c ÷ a = 4 ÷ 5 = 0.80. The length of the major axis is 2a = 10. The latus rectum (the chord through a focus, perpendicular to the major axis) is 2b² ÷ a = 3.60.
Problem 3: focus and directrix of y² = 12x
Compare with y² = 4ax: 4a = 12, so a = 3. The focus is (3, 0) and the directrix is x = -3. The latus rectum has length 4a = 12.
Check with the point where x = 3: y² = 12 × 3, so y = ±6. Its distance to the focus is 6 (straight up), and its distance to the directrix x = -3 is 3 + 3 = 6. Both are 6, exactly as the definition PF = Pd demands.
Problem 4: asymptotes and eccentricity of x²/9 − y²/16 = 1
Here a² = 9, b² = 16, so a = 3 and b = 4. For a hyperbola c² = a² + b² = 9 + 16 = 25, so c = 5.
Foci: (±5, 0). Vertices: (±3, 0). Eccentricity: e = c ÷ a = 5 ÷ 3 = 1.67, greater than 1 as it must be. Asymptotes: y = ±(b/a)x = ±1.33x.
Problem 5: name each curve
4x² + 9y² = 36: both squared, same sign, different coefficients. Divide by 36 to get x²/9 + y²/4 = 1: an ellipse. The discriminant check with A = 4, B = 0, C = 9 gives B² − 4AC = -144, negative.
x² − y² = 4: opposite signs, a hyperbola. With a = b it is called a rectangular hyperbola, and its eccentricity is √2 = 1.414. y² = 8x: only y is squared, so a parabola with 4a = 8, a = 2. x² + y² = 25: a circle of radius 5.
Problem 6: find the ellipse with foci at (±3, 0) and e = 0.60
Since e = c ÷ a and c = 3, a = c ÷ e = 3 ÷ 0.60 = 5. Then b² = a² − c² = 25 − 9 = 16.
The ellipse is x²/25 + y²/16 = 1.
Problem 7: where does the receiver go in a satellite dish?
A dish is a paraboloid; its cross-section is a parabola y² = 4ax. Suppose the dish is 2.4 m across and 0.45 m deep. Put the vertex at the origin: the rim is at x = 0.45, y = 1.2. Substitute: 1.2² = 4a × 0.45, so 1.44 = 1.80a and a = 0.80 m.
The receiver belongs at the focus, 0.80 m from the vertex along the axis. Every ray arriving parallel to the axis reflects through that single point.
Common mistakes
These cost marks every year.
- Mixing up a, b and c. In an ellipse a is the larger one and c² = a² − b². In a hyperbola c² = a² + b², and a is not always the larger of a and b.
- Saying a parabola has e between 0 and 1. It is exactly 1.
- Thinking the plane height changes the shape. It changes the size only; the tilt picks the curve.
- Forgetting that a hyperbola has two branches and a parabola only one.
- Writing the focus of y² = 4ax as (4a, 0). It is (a, 0).
- Using the discriminant of Ax² + Bxy + Cy² on an equation that has already been simplified and dropped the xy term. When B = 0, the sign of AC does the same job.
Where conics show up in real life
The conics are not just exam curves. They describe motion and shape all around you.
- Planets and comets travel on ellipses with the Sun at one focus. A comet that never returns follows a parabola or a hyperbola.
- Satellite dishes, torch reflectors and car headlights are parabolic: they turn a point source into a parallel beam or the reverse.
- Whispering galleries use ellipses: sound from one focus bounces to the other.
- Hyperbolas are used in navigation systems that locate a position from the difference of two signal arrival times.
- A ball thrown in the air follows a parabola (ignoring air).
- The shadow of a lamp shade on a wall is a pair of hyperbola branches.
Exam corner
Students sitting HSC, A-level, JEE or the SAT often get conic questions in these forms.
- Identify the curve from its equation (use the discriminant).
- Find foci, vertices, eccentricity and directrix from the standard equation.
- Write the equation from two facts, such as foci and e, or vertex and focus.
- Find the equation of a tangent or the points where a line meets a conic by solving the two equations together.
- Word problems about dishes, arches and orbits: set up the standard equation and substitute one known point.
Revision summary
If you only remember one thing, remember that a tilt of the cutting plane gives the curve, and e measures it.
Circle: x² + y² = r² (e = 0)
Ellipse: x²/a² + y²/b² = 1, c² = a² − b², e = c/a < 1
Parabola: y² = 4ax, focus (a, 0), directrix x = −a, e = 1
Hyperbola: x²/a² − y²/b² = 1, c² = a² + b², e = c/a > 1
Frequently asked questions
What are conic sections?
Conic sections are the curves you get by slicing a double cone with a plane: the circle, ellipse, parabola and hyperbola. Which one you get depends only on the angle between the plane and the cone’s axis.
Why is a circle a special ellipse?
An ellipse has two foci. When the two foci merge into one point, the sum of distances becomes a constant distance from that centre, and the ellipse is a circle. Its eccentricity is zero.
What is eccentricity?
Eccentricity e is the ratio of a point’s distance from the focus to its distance from the directrix. It measures how far a conic is from a circle: e = 0 circle, 0 < e < 1 ellipse, e = 1 parabola, e > 1 hyperbola.
How do I tell a parabola from a hyperbola in an equation?
A parabola has only one variable squared (y² = 4ax). A hyperbola has both squared with opposite signs (x²/a² − y²/b² = 1). In the general form, the discriminant B² − 4AC is zero for a parabola and positive for a hyperbola.
Where is the focus of a parabola?
For y² = 4ax the focus is at (a, 0), a distance a from the vertex, and the directrix is the line x = −a. Every point of the parabola is equally far from the two.
Why does a hyperbola have two branches?
Because a double cone has two halves joined at the apex. A plane steeper than the cone’s side cuts both halves, and each half contributes one branch.
What happens if the plane passes through the apex of the cone?
That is a degenerate case. The plane cuts the cone in a single point, one straight line, or two crossing lines instead of a proper curve. The simulation keeps the plane away from the apex so you always see a real curve.
Keep studying this topic
The animation made the idea click; now turn it into marks. Syllabus, suggestions, textbooks and admission-test guides are below.
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