Answer key · Unit 11

Conic sections.

Each lesson’s practice questions, their correct answers and the mistakes they spot. Answers read off the Scene use the same setting the learner sees.

Misconception catalogue

Answer key 3

Lesson 1 of 3 · About 20 minutes

Slicing a cone

Four curves from one cut: circle, ellipse, parabola and hyperbola, set by the cut’s tilt against the cone’s side, the radius of a level cut, why the distance from the tip changes only the size, and what happens in a cut through the tip.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
Four curves
Practice · Predict, then look
Question
The cone’s side slopes at $60^\circ$ ⁠ to the level, because it opens $30^\circ$ ⁠ either side of its axis. What will the cut be when Tilt of the cut is $60^\circ$ ⁠, parallel to the side?

Correct answer
- Correct A parabola
Setup, choices and feedback
The Scene at this step
Tilt of the cut
30 °fixed
Distance from the tip
1.8fixed
Show the focus spheres
Offfixed
Choices
- Correct A parabola
- Wrong A very long ellipse
  Misconception: Expects a cut parallel to the cone’s side to close into an ellipse.
  Feedback from the catalogue
  An ellipse closes because the cut crosses every line of the cone in the same half. When the cut is parallel to one of those lines, that line never meets it, and the curve opens out for ever: a parabola. It runs to the rim of the cone and beyond.
- Wrong A hyperbola
  Feedback
  A hyperbola comes when the cut is steeper than the side, cutting both halves of the cone. Parallel to the side it cuts only one.
- Wrong A circle
  Feedback
  A circle comes only from a level cut, at a tilt of $0^\circ$ ⁠.
Once the prediction is locked in, the Scene changes
Tilt of the cut
60 °fixed
The Scene then shows: 1 · E = sin(tilt) / cos(half-angle): 0 circle, below 1 ellipse, 1 parabola, above 1 hyperbola/nodes/cone/outputs/eccentricity
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
A parabola: the cut is parallel to one line of the cone, so it never meets it, and the curve opens out and never closes. This is the same parabola as a quadratic function’s.

Worked solution
1. Less steep than the side ( $\theta < 60^\circ$ ⁠): the cut crosses every line of the cone in one half, and closes: an ellipse.
2. Exactly parallel to the side ( $\theta = 60^\circ$ ⁠): one line of the cone is never cut, and the curve is open: a parabola.
3. Steeper ( $\theta > 60^\circ$ ⁠): the cut crosses the other half too: a hyperbola, with two branches.
Step 2 of 5
Open this step
Hyperbola
Practice · Reach the target
Question
Change Tilt of the cut so that the cut crosses both halves of the cone, then press Check.

Goal
- 3 ± 0.1
  At the start of the step: 1
Setup, choices and feedback
The Scene at this step
Tilt of the cut
30 °
Distance from the tip
1.8fixed
Show the focus spheres
Offfixed
Read from the Scene at
0 circle, 1 ellipse, 2 parabola, 3 hyperbola; through the tip 10 a point, 11 one line, 12 two crossing lines/nodes/cone/outputs/kind
One setting that reaches it
Tilt of the cut
61 °
Feedback for a right answer
Any tilt above $60^\circ$ ⁠, steeper than the cone’s side, gives a hyperbola: one branch in the upper cone and one in the lower.

Feedback for a wrong answer
Not yet. The cut must be steeper than the cone’s side.

Worked solution
1. The cone’s side slopes at $90^\circ - 30^\circ = 60^\circ$ ⁠.
2. A steeper cut, $\theta > 60^\circ$ ⁠, carries on past the tip’s level and crosses the other half too.
3. Set Tilt of the cut to, say, $75^\circ$ ⁠.
Step 3 of 5
Open this step
Circle
Practice · Calculate
Question
The cut is level, $2$ ⁠ units above the tip, and the cone’s side opens $30^\circ$ ⁠ from the axis. What is the circle’s radius? ( $\tan 30^\circ \approx 0.577$ ⁠)

Correct answer
1.155 ± 0.01
Setup, choices and feedback
The Scene at this step
Tilt of the cut
0 °fixed
Distance from the tip
2fixed
Show the focus spheres
Offfixed
Accepted within
± 0.01
Computed from the Scene at this step’s setting
A, half the distance between the vertices (a circle's radius); NaN for a parabola or through the tip/nodes/cone/outputs/semiMajor
Wrong answers it recognises
- 1 ± 0.01
  Misconception: Uses sin or cos of the half-angle for the radius of a level cut, instead of tan.
  Feedback from the catalogue
  In a level cut at height $d$ ⁠ above the tip, the radius $r$ ⁠ and the height $d$ ⁠ are the two legs of a right triangle, and the angle at the tip is the half-angle $\alpha$ ⁠. The opposite leg over the adjacent one is a tangent: $r = d\tan\alpha$ ⁠.
- 1.732 ± 0.01
  Misconception: Uses sin or cos of the half-angle for the radius of a level cut, instead of tan.
  Feedback from the catalogue
  In a level cut at height $d$ ⁠ above the tip, the radius $r$ ⁠ and the height $d$ ⁠ are the two legs of a right triangle, and the angle at the tip is the half-angle $\alpha$ ⁠. The opposite leg over the adjacent one is a tangent: $r = d\tan\alpha$ ⁠.
Feedback for a right answer
$r = 2\tan 30^\circ \approx 1.155$ ⁠. The radius, the height and the cone’s side make a right triangle.

Worked solution
1. Slice the cone along its axis: the height $d = 2$ ⁠, the radius $r$ ⁠ and the cone’s side make a right triangle.
2. The angle at the tip is $30^\circ$ ⁠; $r$ ⁠ is opposite it and $d$ ⁠ adjacent.
3. $\tan 30^\circ = \frac{r}{d}$ ⁠, so $r = 2 \cdot 0.577 \approx 1.155$ ⁠.
If you got $1$ ⁠, you used the sine: $2$ ⁠ is a leg, not the hypotenuse.
Step 4 of 5
Open this step
Size and shape
Practice · Choose
Question
The tilt stays at $30^\circ$ ⁠. What happens to the ellipse as the cut moves further from the tip? Try Distance from the tip, then answer.

Correct answer
- Correct It grows and keeps the same shape
Setup, choices and feedback
The Scene at this step
Tilt of the cut
30 °fixed
Distance from the tip
1
Show the focus spheres
Offfixed
Choices
Shown in a random order.
- Correct It grows and keeps the same shape
- Wrong It becomes rounder
  Misconception: Thinks moving the cut along the cone changes the curve’s shape, not just its size.
  Feedback from the catalogue
  A cone looks the same at every size: moving the cut further from the tip, at the same tilt, is like enlarging the whole picture. Every length grows in the same proportion, so the shape and the eccentricity do not change. Only the tilt sets the shape.
- Wrong It stretches out and gets thinner
  Misconception: Thinks moving the cut along the cone changes the curve’s shape, not just its size.
  Feedback from the catalogue
  A cone looks the same at every size: moving the cut further from the tip, at the same tilt, is like enlarging the whole picture. Every length grows in the same proportion, so the shape and the eccentricity do not change. Only the tilt sets the shape.
- Wrong In the end it opens into a parabola
  Feedback
  The shape depends only on the tilt, and $30^\circ$ ⁠ is less steep than the side: the ellipse stays closed at every distance.
Feedback for a right answer
It grows in the same proportion every way: a cone enlarged looks just like the original. The shape, and the eccentricity too, depend only on the tilt.

Worked solution
1. A cone is similar to itself: enlarging it $k$ ⁠ times about the tip maps it onto itself.
2. The same enlargement takes a cut at distance $d$ ⁠ to a parallel cut at distance $kd$ ⁠.
3. So every length of the cut grows $k$ ⁠ times, and the shape does not change.
Step 5 of 5
Open this step
Through the tip
Practice · Predict, then look
Question
The tilt is $80^\circ$ ⁠, and the cut is a hyperbola. What will the cut be when Distance from the tip is $0$ ⁠, through the tip itself?

Correct answer
- Correct Two straight lines crossing at the tip
Setup, choices and feedback
The Scene at this step
Tilt of the cut
80 °fixed
Distance from the tip
1fixed
Show the focus spheres
Offfixed
Choices
- Correct Two straight lines crossing at the tip
- Wrong A single point
  Feedback
  A point comes when a cut less steep than the side passes through the tip: it touches the cone only there. A steep cut slices through it.
- Wrong One straight line
  Feedback
  One line comes from a cut through the tip parallel to the side, at $60^\circ$ ⁠: it touches the cone along one line.
- Wrong A small hyperbola
  Feedback
  The hyperbola shrinks as the distance does, and at $0$ ⁠ it degenerates: its branches close onto their asymptotes, two straight lines.
Once the prediction is locked in, the Scene changes
Distance from the tip
0fixed
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
Two crossing lines: the cut contains two lines of the cone. These are the degenerate conic sections: a point, one line or two lines, depending on the tilt.

Worked solution
1. A cut through the tip contains the tip, and every line of the cone that lies in it.
2. A cut steeper than the side slices through the cone, so two of the cone’s lines lie in it.
3. The result: two lines crossing at the tip, the asymptotes of the nearby hyperbolas.

Lesson 2 of 3 · About 20 minutes

Eccentricity

One number for a conic section’s shape: the eccentricity e, from 0 for a circle to more than 1 for a hyperbola, how to work it out from the tilt, why size does not change it, and its link to the foci, e = c/a.

Open the lesson

The 5 steps and their answers

Step 1 of 5
Open this step
One number for the shape
Practice · Calculate
Question
Tilt of the cut is $30^\circ$ ⁠, and the cone’s side opens $\alpha = 30^\circ$ ⁠ from the axis. What is the eccentricity, $e = \frac{\sin\theta}{\cos\alpha}$ ⁠? Answer, then choose On under Show the focus spheres.

Correct answer
0.5774 ± 0.005
Setup, choices and feedback
The Scene at this step
Tilt of the cut
30 °fixed
Distance from the tip
1.5fixed
Show the focus spheres
Off
Accepted within
± 0.005
Computed from the Scene at this step’s setting
E = sin(tilt) / cos(half-angle): 0 circle, below 1 ellipse, 1 parabola, above 1 hyperbola/nodes/cone/outputs/eccentricity
Wrong answers it recognises
- 0.5 ± 0.005
  Misconception: Takes the eccentricity as sin of the tilt, forgetting to divide by cos of the half-angle.
  Feedback from the catalogue
  $e = \frac{\sin\theta}{\cos\alpha}$ ⁠: the cut’s tilt is measured against the tilt of the cone’s side. With $\alpha = 30^\circ$ ⁠, $\cos\alpha \approx 0.866$ ⁠, so $e$ ⁠ reaches $1$ ⁠, a parabola, already at $\theta = 60^\circ$ ⁠, when the cut is parallel to the side.
Feedback for a right answer
$e = \frac{0.5}{0.866} \approx 0.577$ ⁠: an ellipse, since $e < 1$ ⁠.

Worked solution
1. $\sin 30^\circ = 0.5$ ⁠ and $\cos 30^\circ \approx 0.866$ ⁠.
2. $e = \frac{0.5}{0.866} \approx 0.577$ ⁠.
If you answered $0.5$ ⁠, you forgot to divide by $\cos\alpha$ ⁠.
Step 2 of 5
Open this step
Setting e
Practice · Reach the target
Question
Change Tilt of the cut so that the eccentricity is $0.5$ ⁠, to within $0.01$ ⁠, then press Check.

Goal
- 0.5 ± 0.01
  At the start of the step: 0.5774
Setup, choices and feedback
The Scene at this step
Tilt of the cut
30 °
Distance from the tip
1.5fixed
Show the focus spheres
Onfixed
Read from the Scene at
E = sin(tilt) / cos(half-angle): 0 circle, below 1 ellipse, 1 parabola, above 1 hyperbola/nodes/cone/outputs/eccentricity
One setting that reaches it
Tilt of the cut
26 °
Feedback for a right answer
$\sin\theta = 0.5 \cdot \cos 30^\circ \approx 0.433$ ⁠, so $\theta \approx 25.7^\circ$ ⁠; the slider gets to $26^\circ$ ⁠, $e \approx 0.506$ ⁠.

Feedback for a wrong answer
Not yet. Find $\sin\theta$ ⁠ from the formula, then $\theta$ ⁠.

Worked solution
1. $e = \frac{\sin\theta}{\cos\alpha} = 0.5$ ⁠, so $\sin\theta = 0.5 \cdot 0.866 = 0.433$ ⁠.
2. $\theta = \sin^{-1} 0.433 \approx 25.7^\circ$ ⁠.
3. Set Tilt of the cut to $26^\circ$ ⁠.
Step 3 of 5
Open this step
Size
Practice · Predict, then look
Question
Distance from the tip is $1.5$ ⁠, and the tilt $30^\circ$ ⁠. What will happen to the eccentricity when the distance doubles to $3$ ⁠?

Correct answer
- Correct It will not change
Setup, choices and feedback
The Scene at this step
Tilt of the cut
30 °fixed
Distance from the tip
1.5fixed
Show the focus spheres
Onfixed
Choices
- Correct It will not change
- Wrong It will double
  Misconception: Thinks moving the cut along the cone changes the curve’s shape, not just its size.
  Feedback from the catalogue
  A cone looks the same at every size: moving the cut further from the tip, at the same tilt, is like enlarging the whole picture. Every length grows in the same proportion, so the shape and the eccentricity do not change. Only the tilt sets the shape.
- Wrong It will halve
  Misconception: Thinks moving the cut along the cone changes the curve’s shape, not just its size.
  Feedback from the catalogue
  A cone looks the same at every size: moving the cut further from the tip, at the same tilt, is like enlarging the whole picture. Every length grows in the same proportion, so the shape and the eccentricity do not change. Only the tilt sets the shape.
Once the prediction is locked in, the Scene changes
Distance from the tip
3fixed
Show the focus spheres
Onfixed
The Scene then shows: 0.5774 · E = sin(tilt) / cos(half-angle): 0 circle, below 1 ellipse, 1 parabola, above 1 hyperbola/nodes/cone/outputs/eccentricity
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
It stays at $0.577$ ⁠. $a$ ⁠ and $c$ ⁠ both double, and the ratio $e = \frac{c}{a}$ ⁠ does not change: eccentricity describes shape, not size.

Worked solution
1. $e = \frac{\sin\theta}{\cos\alpha}$ ⁠ depends only on the tilt and the cone’s half-angle.
2. The distance does not appear in it.
3. So $e$ ⁠ stays at $0.577$ ⁠, and the ellipse only grows.
Step 4 of 5
Open this step
The foci
Practice · Calculate
Question
In this ellipse the semi-major axis is $a \approx 1.30$ ⁠, and the eccentricity $e \approx 0.577$ ⁠. What is $c$ ⁠, the distance from the centre to each focus?

Correct answer
0.75 ± 0.01
Setup, choices and feedback
The Scene at this step
Tilt of the cut
30 °fixed
Distance from the tip
1.5fixed
Show the focus spheres
Onfixed
Accepted within
± 0.01
Computed from the Scene at this step’s setting
C, from the centre to each focus, where the Dandelin spheres touch the cut; NaN for a parabola or through the tip/nodes/cone/outputs/focalDistance
Wrong answers it recognises
- 2.25 ± 0.02
  Misconception: Turns e = c/a upside down, as a/c.
  Feedback from the catalogue
  $e = \frac{c}{a}$ ⁠: the distance from the centre to a focus over the semi-major axis. In an ellipse the foci are inside it, $c < a$ ⁠, so $e < 1$ ⁠. If your $c$ ⁠ came out bigger than $a$ ⁠, the focus would be outside the ellipse.
Feedback for a right answer
$c = e \cdot a \approx 0.577 \cdot 1.30 = 0.75$ ⁠. The foci are inside the ellipse, nearer the ends than the centre.

Worked solution
1. $e = \frac{c}{a}$ ⁠, so $c = e \cdot a$ ⁠.
2. $c \approx 0.577 \cdot 1.299 \approx 0.75$ ⁠.
If you got $2.25$ ⁠, you worked out $\frac{a}{e}$ ⁠: the focus would be outside the ellipse.
Step 5 of 5
Open this step
When e is 1
Practice · Choose
Question
At what Tilt of the cut is the eccentricity exactly $1$ ⁠, in a cone that opens $30^\circ$ ⁠ from its axis?

Correct answer
- Correct $60^\circ$ ⁠
Setup, choices and feedback
The Scene at this step
Tilt of the cut
45 °
Distance from the tip
1.5fixed
Show the focus spheres
On
Choices
Shown in a random order.
- Correct $60^\circ$ ⁠
- Wrong $30^\circ$ ⁠
  Misconception: Takes the eccentricity as sin of the tilt, forgetting to divide by cos of the half-angle.
  Feedback from the catalogue
  $e = \frac{\sin\theta}{\cos\alpha}$ ⁠: the cut’s tilt is measured against the tilt of the cone’s side. With $\alpha = 30^\circ$ ⁠, $\cos\alpha \approx 0.866$ ⁠, so $e$ ⁠ reaches $1$ ⁠, a parabola, already at $\theta = 60^\circ$ ⁠, when the cut is parallel to the side.
- Wrong $45^\circ$ ⁠
  Feedback
  $e = \frac{\sin 45^\circ}{\cos 30^\circ} \approx 0.82$ ⁠: still an ellipse.
- Wrong $90^\circ$ ⁠, an upright cut
  Feedback
  An upright cut is steeper than the side: $e = \frac{1}{0.866} \approx 1.15$ ⁠, a hyperbola.
Feedback for a right answer
$60^\circ$ ⁠: $\sin 60^\circ = \cos 30^\circ$ ⁠, so $e = 1$ ⁠, a parabola. It is exactly the slope of the cone’s side.

Worked solution
1. $e = 1$ ⁠ when $\sin\theta = \cos\alpha = \cos 30^\circ$ ⁠.
2. $\cos 30^\circ = \sin 60^\circ$ ⁠, so $\theta = 60^\circ$ ⁠.
3. $60^\circ = 90^\circ - \alpha$ ⁠: the cut is parallel to the cone’s side.

Lesson 3 of 3 · About 25 minutes

Foci and Dandelin spheres

Why an ellipse has two foci whose distances add up to a constant: Dandelin spheres, the gardener’s ellipse with a string, the link between a, b and c, the parabola with one focus and a directrix, and the hyperbola with a fixed difference.

Open the lesson

The 6 steps and their answers

Step 1 of 6
Open this step
The gardener’s ellipse
Practice · Calculate
Question
The ellipse is $3.12$ ⁠ units long from vertex to vertex, so $2a \approx 3.12$ ⁠. $P$ ⁠ is any point on the ellipse. What is $PF_1 + PF_2$ ⁠, the sum of the distances from $P$ ⁠ to the two foci?

Correct answer
3.118 ± 0.02
Setup, choices and feedback
The Scene at this step
Tilt of the cut
30 °fixed
Distance from the tip
1.8fixed
Show the focus spheres
Onfixed
Accepted within
± 0.02
Computed from the Scene at this step’s setting
A, half the distance between the vertices (a circle's radius); NaN for a parabola or through the tip/nodes/cone/outputs/semiMajor
Wrong answers it recognises
- 1.56 ± 0.02
  Misconception: Takes the sum of the distances to the two foci as a, half the major axis.
  Feedback from the catalogue
  At a vertex of the ellipse, $P$ ⁠ is $a - c$ ⁠ from one focus and $a + c$ ⁠ from the other, which together make $2a$ ⁠: the whole major axis, vertex to vertex. That sum is the same at every point of the ellipse.
Feedback for a right answer
$PF_1 + PF_2 = 2a \approx 3.12$ ⁠, at every point of the ellipse. You can draw an ellipse with a string $2a$ ⁠ long whose ends are tied at the foci.

Worked solution
1. The focus spheres touch the cone along two circles, and the cut at the foci.
2. From $P$ ⁠, two tangents to the same sphere are equal: $PF_1$ ⁠ equals the distance from $P$ ⁠ along the cone’s side to the first sphere’s circle, and $PF_2$ ⁠ to the second’s.
3. So $PF_1 + PF_2$ ⁠ is the distance along the cone’s side between the two circles, the same for every $P$ ⁠.
4. At a vertex the sum is $(a - c) + (a + c) = 2a \approx 3.12$ ⁠.
If you answered $1.56$ ⁠, that is only $a$ ⁠.
Step 2 of 6
Open this step
a, b and c
Practice · Calculate
Question
In the same ellipse $a \approx 1.559$ ⁠ and $c = 0.9$ ⁠. What is $b$ ⁠, the semi-minor axis?

Correct answer
1.273 ± 0.01
Setup, choices and feedback
The Scene at this step
Tilt of the cut
30 °fixed
Distance from the tip
1.8fixed
Show the focus spheres
Onfixed
Accepted within
± 0.01
Computed from the Scene at this step’s setting
B, with a² = b² + c² (ellipse) or c² = a² + b² (hyperbola); NaN for a parabola or through the tip/nodes/cone/outputs/semiMinor
Wrong answers it recognises
- 1.8 ± 0.01
  Misconception: Uses b² = a² + c² for an ellipse, which is the hyperbola’s rule.
  Feedback from the catalogue
  In an ellipse $a^2 = b^2 + c^2$ ⁠: from the end of the minor axis, the distance to each focus is $a$ ⁠, the hypotenuse of a triangle with legs $b$ ⁠ and $c$ ⁠. So $b = \sqrt{a^2 - c^2}$ ⁠, smaller than $a$ ⁠. In a hyperbola it is the other way round: $c^2 = a^2 + b^2$ ⁠.
Feedback for a right answer
$b = \sqrt{a^2 - c^2} = \sqrt{2.43 - 0.81} \approx 1.27$ ⁠.

Worked solution
1. From the end of the minor axis, $PF_1 = PF_2$ ⁠, and they add up to $2a$ ⁠: each is $a$ ⁠.
2. $a$ ⁠ is the hypotenuse of a triangle with legs $b$ ⁠ and $c$ ⁠: $a^2 = b^2 + c^2$ ⁠.
3. $b = \sqrt{1.559^2 - 0.9^2} = \sqrt{2.430 - 0.810} = \sqrt{1.620} \approx 1.27$ ⁠.
If you got $1.8$ ⁠, you added the squares: that is the hyperbola’s rule.
Step 3 of 6
Open this step
One focus
Practice · Predict, then look
Question
The tilt is $50^\circ$ ⁠, and both focus spheres are shown. What will happen to the larger ball when the tilt reaches $60^\circ$ ⁠ and the cut is a parabola?

Correct answer
- Correct It will grow and move off without end: a parabola has one focus
Setup, choices and feedback
The Scene at this step
Tilt of the cut
50 °
Distance from the tip
1.8fixed
Show the focus spheres
Onfixed
Choices
- Correct It will grow and move off without end: a parabola has one focus
- Wrong It will shrink until it vanishes
  Feedback
  The larger ball sits beyond the cut, where the cone is wide. The closer the cut comes to parallel with the side, the further up the cone it is pushed, and the bigger it grows.
- Wrong It will merge with the small ball
  Feedback
  The two balls merge only for a level cut, when the ellipse is a circle and both foci are at its centre.
Once the prediction is locked in, the Scene changes
Tilt of the cut
60 °
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
The second ball is pushed off to infinity, and the second focus with it: a parabola has one focus and one directrix. You can think of it as an ellipse whose second focus is infinitely far away.

Worked solution
1. The larger ball’s centre is on the axis, $\frac{d}{\cos\theta - \sin\alpha}$ ⁠ from the tip.
2. As $\theta$ ⁠ approaches $60^\circ$ ⁠, $\cos\theta$ ⁠ approaches $\sin 30^\circ = 0.5$ ⁠, and the denominator approaches zero.
3. So the distance, and the ball, grow without limit, leaving only the small ball, with the parabola’s single focus.
Step 4 of 6
Open this step
The directrix
Practice · Choose
Question
The point running along the parabola shows its distance to the focus, $PF$ ⁠, and to the directrix, $PD$ ⁠. The eccentricity is $e = \frac{PF}{PD} = 1$ ⁠. What is true at every point of the parabola?

Correct answer
- Correct $PF = PD$ ⁠
Setup, choices and feedback
The Scene at this step
Tilt of the cut
60 °fixed
Distance from the tip
1.8fixed
Show the focus spheres
Onfixed
Choices
Shown in a random order.
- Correct $PF = PD$ ⁠
- Wrong $PF + PD$ ⁠ is fixed
  Feedback
  A fixed sum of distances belongs to the ellipse, and there both distances are to foci. On a parabola the distance to the focus and the distance to the directrix grow together, so their sum is not fixed.
- Wrong $PF = 2 \cdot PD$ ⁠
  Feedback
  $PF = 2 \cdot PD$ ⁠ means $e = 2$ ⁠: a hyperbola, not a parabola.
Feedback for a right answer
$PF = PD$ ⁠: every point on the parabola is exactly as far from the focus as from the directrix. That is the parabola’s definition in the plane, and it gives the quadratic function: a focus at $(0, p)$ ⁠ and the directrix $y = -p$ ⁠ give $y = \frac{x^2}{4p}$ ⁠.

Worked solution
1. On every conic section $\frac{PF}{PD} = e$ ⁠.
2. On a parabola $e = 1$ ⁠, so $PF = PD$ ⁠.
3. In the plane: $PF^2 = x^2 + (y - p)^2$ ⁠ and $PD^2 = (y + p)^2$ ⁠. Setting them equal gives $x^2 = 4py$ ⁠.
Step 5 of 6
Open this step
Hyperbola
Practice · Choose
Question
The tilt is $80^\circ$ ⁠: a hyperbola, with a focus inside each branch. $P$ ⁠ is on one branch. What stays the same along the hyperbola?

Correct answer
- Correct The difference $|PF_1 - PF_2|$ ⁠
Setup, choices and feedback
The Scene at this step
Tilt of the cut
80 °fixed
Distance from the tip
1.8
Show the focus spheres
Onfixed
Choices
Shown in a random order.
- Correct The difference $|PF_1 - PF_2|$ ⁠
- Wrong The sum $PF_1 + PF_2$ ⁠
  Misconception: Expects the distances to a hyperbola’s foci to have a fixed sum, as an ellipse’s do, rather than a fixed difference.
  Feedback from the catalogue
  On a hyperbola each branch is further from one focus than from the other, and at every point the difference of the distances is $2a$ ⁠. The sum grows without limit as you go out along a branch.
- Wrong The product $PF_1 \cdot PF_2$ ⁠
  Feedback
  As $P$ ⁠ moves out along the branch, both distances grow, and so does their product.
Feedback for a right answer
The difference is fixed, $2a$ ⁠. The spheres now sit on either side of the tip, one in each half of the cone, so the tangents from $P$ ⁠ to them run opposite ways along the cone’s side, and it is their difference that stays fixed.

Worked solution
1. On a hyperbola the spheres are in the two halves of the cone, either side of the tip.
2. $PF_1$ ⁠ and $PF_2$ ⁠ equal the distances from $P$ ⁠ along the cone’s side to the two circles.
3. Both circles lie the same way from $P$ ⁠ along that line, so their difference is the fixed distance between the circles: $|PF_1 - PF_2| = 2a$ ⁠.
4. At a vertex: $(c + a) - (c - a) = 2a$ ⁠.
Step 6 of 6
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Vertex and focus
Practice · Calculate
Question
In the ellipse $a \approx 1.559$ ⁠, and the distance from the centre to each focus is $c = 0.9$ ⁠. What is the distance from a vertex to the focus nearest it?

Correct answer
0.6588 ± 0.01
Setup, choices and feedback
The Scene at this step
Tilt of the cut
30 °fixed
Distance from the tip
1.8fixed
Show the focus spheres
Onfixed
Accepted within
± 0.01
Computed from the Scene at this step’s setting
From a vertex to the focus beside it: a − c, the parabola's focal length, or c − a; NaN through the tip/nodes/cone/outputs/vertexToFocus
Wrong answers it recognises
- 2.459 ± 0.01
  Misconception: Adds the centre-to-focus distance to a instead of subtracting it, giving the far focus.
  Feedback from the catalogue
  A vertex and a focus on the same side of the centre are close together: the distance between them is $a - c$ ⁠. The distance $a + c$ ⁠ is to the focus on the other side of the centre.
Feedback for a right answer
$a - c \approx 0.659$ ⁠. To the other focus $a + c$ ⁠, together $2a$ ⁠.

Hint
The vertex is $a$ ⁠ from the centre, and the focus $c$ ⁠ from the centre, in the same direction.

Worked solution
1. The vertex and the near focus are on the same side of the centre.
2. The distance between them: $a - c = 1.559 - 0.9 \approx 0.659$ ⁠.
3. To the far focus: $a + c \approx 2.459$ ⁠, together $2a$ ⁠, as for every point on the ellipse.
In a planet’s orbit, $a - c$ ⁠ is the nearest distance to the Sun, and $a + c$ ⁠ the farthest.

Across the unit · 9 misconceptions

Misconception catalogue

Each common mistake this unit recognises, in a few words. Open one for what the learner reads when an answer matches it, and where it is asked.

Index

Slicing a cone

Four curves

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Hyperbola

Question

Goal

The Scene at this step

One setting that reaches it

Worked solution

Circle

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Worked solution

Size and shape

Question

Correct answer

The Scene at this step

Choices

Worked solution

Through the tip

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Eccentricity

One number for the shape

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Worked solution

Setting e

Question

Goal

The Scene at this step

One setting that reaches it

Worked solution

Size

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

The foci

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Worked solution

When e is 1

Question

Correct answer

The Scene at this step

Choices

Worked solution

Foci and Dandelin spheres

The gardener’s ellipse

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Worked solution

a, b and c

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Worked solution

One focus

Question