Precalculus

Conic sections from a double cone

A flashlight beam on a wall can make a round spot, a stretched oval, or a patch that never closes. What decides which one you get?

You need

Cross sections of solidsMath 7

Unlocks

Focus and directrixPrecalculus

Where you are 6 in Conic sections

Geometry11

Conic sections6

Conic sections from a double cone

Focus and directrix The ellipse The hyperbola The parabola as a locus Identifying and translating conics

Why this exists

A circle closes up and is round. An ellipse closes up and is stretched. A parabola is one open curve, and a hyperbola is two open pieces. They look like four unrelated curves, yet Greek geometers, Apollonius of Perga above all, studied all four as cuts through one solid, long before coordinates existed.

In cross sections of solids, a cone cut straight across gave a circle and a cut on a slant gave an ellipse. This lesson finishes that story. Cut a cone that runs on without end in both directions, and only the tilt of the plane changes from one curve to the next. One comparison of two angles decides the curve, which is why the four are one family, the conic sections.

Definition

Take a line, the axis, and a second line crossing it at a point V at an angle \alpha, with 0^\circ < \alpha < 90^\circ. Spin the second line all the way around the axis, keeping the angle fixed. The surface it sweeps out is a double cone (a right circular double cone).

V is the vertex.
Each position of the spinning line is a side line. Every side line passes through V and makes the angle \alpha with the axis, the cone’s half-angle.
The two halves that meet at V are the nappes. Both run on without end.

Figure 1A double cone seen from the side. Both nappes run on without end.

A conic section is the set of points where a plane meets the double cone. The plane’s tilt is measured by \beta, the angle between the plane and the axis: the smallest angle the axis makes with any line in the plane, so 0^\circ \le \beta \le 90^\circ. A plane perpendicular to the axis has \beta = 90^\circ; a plane parallel to the axis, or containing it, has \beta = 0^\circ.

Intuition

Look along the cutting plane, from a direction perpendicular to the axis. The plane is seen edge-on as a line, the cone’s outline is two side lines crossing at V, and the cut line meets the axis at \beta, true size. Figure 2 tilts the plane step by step.

Figure 2The same cone cut at four tilts, seen with the plane edge-on.

Perpendicular to the axis. Every point of the cut is the same distance from V along the axis, so it is the same distance from the axis: a circle.

Tilted a little. The curve stretches, but the cut line still crosses both side lines on the same nappe and the curve closes up: an ellipse.

Parallel to a side line. The cut line never meets that side line, so the curve never closes. It stays on one nappe and runs off without end: a parabola.

Steeper than the sides. Now the cut line crosses one side line above V and the other below V. The plane slices both nappes, and the curve comes in two pieces: a hyperbola.

Figure 3The whole classification is the comparison of \beta with \alpha.

Faced straight on, the four cuts look like Figure 4.

Figure 4What each cut looks like when you face the plane.

A flashlight beam is one nappe of light and a wall is the plane, which is why the spot changes shape as you tip the flashlight.

Results

The classification. Cut a double cone of half-angle \alpha with a plane at angle \beta to the axis.

Tilt of the plane	Plane misses the vertex	Plane through the vertex
\beta = 90^\circ	circle	a point
\alpha < \beta < 90^\circ	ellipse	a point
\beta = \alpha	parabola	one line
0^\circ \le \beta < \alpha	hyperbola	two lines crossing at V

In words: the plane is flatter than the side lines (\beta > \alpha), parallel to one (\beta = \alpha), or steeper (\beta < \alpha). A circle is the ellipse with \beta = 90^\circ. The three cuts through the vertex are the degenerate conics: a point, a line, and a pair of crossing lines.

The size of a circle. A plane perpendicular to the axis at distance h from V cuts a circle centered on the axis with radius

r = h\tan\alpha

because a side line, the axis and a radius form a right triangle with angle \alpha at V (right triangle trigonometry). With the center as origin in that plane, the circle is x^2 + y^2 = r^2 (the circle’s equation).

Why it is true

The proof counts the side lines that run parallel to the plane.

Let V be the vertex and \vec{n} a unit vector perpendicular to the plane. For each side line, let \vec{u} be the unit vector along it toward the upper nappe, and let k = \vec{u}\cdot\vec{n} (the dot product): how far one unit step along the side line moves you in the direction of \vec{n}. Split \vec{u} into \cos\alpha along the axis and \sin\alpha straight out from it, at some angle \theta around the axis. Since \vec{n} makes the angle 90^\circ - \beta with the axis, it splits into \sin\beta along the axis and \cos\beta straight out, at some angle \varphi. An along-axis part and a straight-out part are perpendicular and give 0; the two straight-out parts are \theta - \varphi apart. So

k(\theta) = \cos\alpha\sin\beta + \sin\alpha\cos\beta\cos(\theta - \varphi)

As \theta goes once around, k swings between a largest value \sin(\alpha + \beta) and a smallest value

\cos\alpha\sin\beta - \sin\alpha\cos\beta = \sin(\beta - \alpha)

by the difference formula for sine. A side line is parallel to the plane exactly when k = 0.

\beta > \alpha: the smallest value is positive, so k is never 0. No side line is parallel to the plane.
\beta = \alpha: the smallest value is 0, reached at one \theta. Exactly one side line is parallel, and k touches 0 there without changing sign.
\beta < \alpha: the smallest value is negative, so k crosses 0 twice. Two side lines are parallel, and k is positive on one arc of side lines and negative on the other.

Now let c be the distance from V to the plane, measured along \vec{n}. A side line meets the plane where t\,k = c, at distance t = c/k from V: on the upper nappe if t > 0, the lower if t < 0.

Plane misses V (c \ne 0). With no parallel side line, t keeps one sign and stays bounded, so every side line gives one point, all on one nappe, and they join up as \theta goes around: a closed curve, the ellipse. When \beta = 90^\circ, k = \cos\alpha for every side line, so every point is at the same distance from V, the circle. With one parallel side line, t keeps one sign but grows without end as \theta nears it: one open curve on one nappe, the parabola. With two, t changes sign: one arc of side lines lands on the upper nappe, the other on the lower, each running off at both ends: the two branches of the hyperbola.

Plane through V (c = 0). Then t\,k = 0. A side line with k \ne 0 meets the plane only at V, and a side line with k = 0 lies in the plane. Zero, one or two such lines give a point, one line, or two crossing lines.

This settles which family each cut belongs to. That each is exactly the curve defined by its foci is the focus and directrix story that comes next.

Worked examples

Example 1

Naming the cut from the two angles

A double cone has half-angle 35^\circ. Name the curve cut by a plane that misses the vertex and makes, with the axis, an angle of (a) 90^\circ, (b) 60^\circ, (c) 35^\circ, (d) 20^\circ, (e) 0^\circ.

Solution

Compare each \beta with \alpha = 35^\circ.

(a) \beta = 90^\circ: the plane is perpendicular to the axis, a circle.

(b) 60^\circ > 35^\circ: flatter than the sides, an ellipse.

(d) 20^\circ < 35^\circ: steeper than the sides, a hyperbola.

(e) 0^\circ < 35^\circ: a plane parallel to the axis is the steepest of all, a hyperbola.

Example 2

How big is the circle

Sadiq cuts a double cone of half-angle 30^\circ with a plane perpendicular to the axis, 6 units from the vertex. Find the circle’s radius and its equation in that plane, centered on the axis.

Solution

The axis, a side line and a radius make a right triangle: the leg along the axis is 6, the angle at V is 30^\circ, and the radius is the leg opposite that angle. So

r = 6\tan 30^\circ = 6 \cdot \frac{\sqrt{3}}{3} = 2\sqrt{3}

and r^2 = 12, so the circle is x^2 + y^2 = 12.

Example 3

A tilted cut in coordinates

In a side view the cone is y = x and y = -x, with the axis along the y-axis and the vertex at the origin. A cutting plane shows as the line \displaystyle y = \frac{1}{2}x + 3. Find where it meets the two side lines, name the curve, and find the curve’s length along the tilt.

Figure 5Examples 3 and 4. Which nappe each meeting point is on decides the curve.

Solution

Meet y = x: \displaystyle x = \frac{1}{2}x + 3, so \displaystyle \frac{1}{2}x = 3 and the point is (6, 6).

Meet y = -x: \displaystyle -x = \frac{1}{2}x + 3, so \displaystyle -\frac{3}{2}x = 3, x = -2, and the point is (-2, 2).

Both points have y > 0, so both are on the upper nappe: the plane cuts one nappe and closes up, an ellipse. The angles agree. Each side line has slope 1, so \alpha = 45^\circ. The cut line rises 1 for every 2 across, so its angle with the vertical axis has \tan\beta = 2 and \beta \approx 63.4^\circ > 45^\circ (inverse tangent).

The two points are the ends of the ellipse in this view, so its length along the tilt is

\begin{aligned} \sqrt{(6 - (-2))^2 + (6 - 2)^2} &= \sqrt{64 + 16} \\ &= \sqrt{80} = 4\sqrt{5} \end{aligned}

Example 4

A steep cut in coordinates

Same cone. The cutting plane shows as y = 3x - 4. Find the meeting points and name the curve.

Solution

Meet y = x: x = 3x - 4, so x = 2, the point (2, 2).

Meet y = -x: -x = 3x - 4, so x = 1, the point (1, -1).

The first point is above the vertex and the second is below it, so the plane cuts both nappes: a hyperbola. In angles, \displaystyle \tan\beta = \frac{1}{3}, so \beta \approx 18.4^\circ < 45^\circ. The two points are the turning points of the two branches, and they are \displaystyle \sqrt{1^2 + 3^2} = \sqrt{10} apart.

Example 5

Planes through the vertex

A double cone has half-angle 25^\circ. Each plane below passes through the vertex. Name the cut when the plane makes, with the axis, (a) 60^\circ, (b) 25^\circ, (c) 0^\circ.

Figure 6A plane through the vertex gives a point, one line, or two crossing lines.

Solution

(a) 60^\circ > 25^\circ: the plane is flatter than every side line, so it meets each one only at V. The cut is a point.

(b) \beta = \alpha: the plane is parallel to one side line and passes through V, which is on it, so it contains that whole line and touches the cone nowhere else. The cut is one line.

(c) 0^\circ < 25^\circ: the plane contains the axis, and in it lie the two side lines at 25^\circ on either side of the axis. The cut is two lines crossing at V at an angle of 2 \cdot 25^\circ = 50^\circ.

Example 6

The flashlight on the wall

Maryam’s flashlight throws a beam with half-angle 20^\circ. (a) She aims it straight at a wall 3 meters away. How wide is the spot? (b) Name the edge of the lit patch when the beam’s axis makes, with the wall, an angle of 50^\circ, 20^\circ and 8^\circ.

Solution

(a) Straight on, the wall is perpendicular to the axis, so the spot is a circle of radius 3\tan 20^\circ \approx 1.09 meters, about 2.18 meters across.

(b) 50^\circ > 20^\circ: an ellipse. At 20^\circ the wall is parallel to one edge of the beam, so the patch never closes: a parabola. At 8^\circ the wall is steeper than the beam’s edge: a hyperbola, but only one branch, because the beam is one nappe.

Example 7

Finding the half-angle first

Kamsi knows that a plane perpendicular to the axis of a double cone, \displaystyle 4\sqrt{3} units from the vertex, cuts a circle of radius 12. Which tilts of a plane that misses the vertex give an ellipse, a parabola, and a hyperbola?

Solution

From r = h\tan\alpha,

\tan\alpha = \frac{12}{4\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}

so \alpha = 60^\circ. Then the cut is an ellipse for 60^\circ < \beta < 90^\circ (a circle at 90^\circ), a parabola at \beta = 60^\circ, and a hyperbola for 0^\circ \le \beta < 60^\circ.

Boundaries

The cone is a right circular double cone that runs on without end. A single nappe gives only one branch of a hyperbola, and a cone that stops at a base can cut a curve short.

The half-angle is strictly between 0^\circ and 90^\circ. At 0^\circ the side lines are parallel to the axis and the solid is a cylinder; at 90^\circ the cone flattens into a plane.

\beta is measured from the axis. Measured from the horizontal, the tilt is 90^\circ - \beta, and the comparison flips.

Through the vertex, the cut is degenerate. A point, a line, or two crossing lines. Two parallel lines never come from a cone; they come from a cylinder.

Where this goes next

Slicing shows why the four curves are one family, but gives no equation. The next lesson, focus and directrix, describes each conic by distances in its own plane, with no cone in sight. From there each curve gets its own lesson: the parabola, the ellipse and the hyperbola. Finally identifying a conic from its equation reads the same four shapes, and the degenerate ones, off the squared terms of an equation.

Exercises

Naming the cut

A double cone has half-angle 40^\circ. Name the cut for a plane at the given angle \beta to the axis.

\beta = 90^\circ, missing the vertex
\beta = 65^\circ, missing the vertex
\beta = 40^\circ, missing the vertex
\beta = 12^\circ, missing the vertex
\beta = 0^\circ, missing the vertex
\beta = 75^\circ, through the vertex
\beta = 40^\circ, through the vertex
\beta = 0^\circ, through the vertex. Give the angle between the lines too.

Circles

Half-angle 45^\circ, plane perpendicular to the axis 5 units from the vertex. Find the radius and the equation, centered on the axis.
Half-angle 60^\circ, plane perpendicular to the axis 3 units from the vertex. Find the radius and the area of the circle.
A plane perpendicular to the axis, \displaystyle 7\sqrt{3} units from the vertex, cuts a circle of radius 7. Find the half-angle.

Side views

In Exercises 12 to 15 the cone is y = x and y = -x, vertex at the origin, axis along the y-axis.

The cut line is \displaystyle y = \frac{1}{3}x + 4. Find the meeting points, name the curve, and find its length along the tilt.
The cut line is y = 2x - 3. Find the meeting points, name the curve, and find the distance between the two points.
The cut line is y = -x + 6. Find every meeting point and name the curve.
Name the cut for the line y = 4x and for the line \displaystyle y = \frac{1}{4}x, each a plane through the vertex.
The cone is \displaystyle y = \sqrt{3}x and \displaystyle y = -\sqrt{3}x. The cut line is y = x + 6. Find the meeting points, name the curve, and find its length along the tilt.

Light on a wall

Maryam’s other flashlight has half-angle 12^\circ. Name the edge of the lit patch when the beam’s axis makes, with the wall, 90^\circ, 30^\circ, 12^\circ and 5^\circ.
She aims it straight at a wall 2.5 meters away. Find the radius of the spot to the nearest hundredth of a meter.
Sadiq’s lamp has a lampshade open at the top and the bottom, so it throws light up and down in a double cone of half-angle 35^\circ with a vertical axis. It stands near a vertical wall. What curve edges the light on the wall, and how many pieces does it have?

Reasoning

Explain, with a side view, why a hyperbola uses both nappes while an ellipse uses one.
Explain why a plane perpendicular to the axis can never cut a parabola or a hyperbola, whatever the cone.
Kamsi says that a plane through the vertex, tilted just right, cuts a tiny ellipse. Explain why it cannot.
Explain why every plane parallel to the axis that misses the vertex cuts a hyperbola, for every double cone.
A cylinder’s side lines are all parallel to its axis. Explain why no plane cuts a cylinder in a parabola or a hyperbola.

Answers

A circle. The plane is perpendicular to the axis.
An ellipse, since 65^\circ > 40^\circ.
A parabola, since \beta = \alpha.
A hyperbola, since 12^\circ < 40^\circ.
A hyperbola, since 0^\circ < 40^\circ; it cuts both nappes.
A point, since 75^\circ > 40^\circ: the plane meets every side line only at the vertex.
One line: the plane contains the one side line it is parallel to.
Two lines crossing at the vertex at 80^\circ. The plane contains the axis and the two side lines at 40^\circ on either side of it, 2 \cdot 40^\circ = 80^\circ.
r = 5\tan 45^\circ = 5, so x^2 + y^2 = 25.
\displaystyle r = 3\tan 60^\circ = 3\sqrt{3}, and the area is \displaystyle \pi(3\sqrt{3})^2 = 27\pi.
30^\circ. \displaystyle \tan\alpha = \frac{7}{7\sqrt{3}} = \frac{1}{\sqrt{3}}.
(6, 6) and (-3, 3), an ellipse of length \displaystyle 3\sqrt{10}. From \displaystyle x = \frac{1}{3}x + 4, x = 6; from \displaystyle -x = \frac{1}{3}x + 4, x = -3. Both points are above the vertex. Length \displaystyle \sqrt{9^2 + 3^2} = \sqrt{90}. Also \tan\beta = 3, \beta \approx 71.6^\circ > 45^\circ.
(3, 3) and (1, -1), a hyperbola; the points are \displaystyle 2\sqrt{5} apart. From x = 2x - 3, x = 3; from -x = 2x - 3, x = 1. One point is above the vertex and one below. Distance \displaystyle \sqrt{2^2 + 4^2} = \sqrt{20}.
Only (3, 3), a parabola on the upper nappe. The line is parallel to y = -x, so it never meets it; x = -x + 6 gives x = 3.
y = 4x: two crossing lines, since \displaystyle \tan\beta = \frac{1}{4} gives \beta \approx 14.0^\circ < 45^\circ. \displaystyle y = \frac{1}{4}x: a point, since \tan\beta = 4 gives \beta \approx 76.0^\circ > 45^\circ.
\displaystyle (3 + 3\sqrt{3},\ 9 + 3\sqrt{3}) and \displaystyle (3 - 3\sqrt{3},\ 9 - 3\sqrt{3}), an ellipse of length \displaystyle 6\sqrt{6}. From \displaystyle \sqrt{3}x = x + 6, \displaystyle x = \frac{6}{\sqrt{3} - 1} = 3 + 3\sqrt{3}; from \displaystyle -\sqrt{3}x = x + 6, \displaystyle x = -\frac{6}{\sqrt{3} + 1} = 3 - 3\sqrt{3}. Both y values are positive. Across and up both differ by \displaystyle 6\sqrt{3}, so the length is \displaystyle 6\sqrt{3}\cdot\sqrt{2}.
A circle, an ellipse, a parabola, and one branch of a hyperbola: 90^\circ is perpendicular, 30^\circ > 12^\circ, 12^\circ = 12^\circ, 5^\circ < 12^\circ.
About 0.53 meters: 2.5\tan 12^\circ \approx 0.531.
A hyperbola in two pieces, one above the lamp and one below. The wall is parallel to the axis, so \beta = 0^\circ < 35^\circ, and the light runs both ways, so both branches appear.
In the side view, a cut line flatter than the sides crosses both side lines above the vertex (or both below), so the curve closes on one nappe. A steeper one crosses one side line above the vertex and the other below, so the curve has a piece on each nappe.
Such a plane has \beta = 90^\circ, and every cone has \alpha < 90^\circ, so \beta > \alpha always: a circle, never a parabola or hyperbola.
Through the vertex, a side line not lying in the plane meets it only at the vertex, so the cut is made of the vertex and whole side lines: a point, one line, or two lines. An ellipse would need points other than the vertex on side lines that are not in the plane.
Such a plane has \beta = 0^\circ, and every cone has \alpha > 0^\circ, so \beta < \alpha: a hyperbola.
A plane parallel to one side line is parallel to all of them, and so to the axis. Such a plane meets the cylinder in two parallel lines, one line, or nothing. Any other plane meets every side line exactly once, in a closed curve, a circle or an ellipse. An open curve needs some side lines parallel to the plane and others not.