Conic sections

The last chapter ended with a promise: to cut a cone with a plane. You can do it right now without leaving the room. Switch on a torch and shine it at the wall. Light leaves the bulb as a spreading cone, and the wall cuts it. Point the torch straight at the wall and the patch is round. Tilt it and the patch stretches into an oval. Keep tilting, and at some point the far edge of the patch runs off to infinity: the boundary of the light becomes an open curve. And in the evening look at the wall next to a standard lamp. The shade sends light up and down in two cones, and on the wall they draw two curves facing opposite ways.

A circle, an oval, an open arc and a pair of arcs: these are the curves the ancient Greeks called conic sections. They were studied out of pure curiosity, with no use at all, for almost two thousand years. And then it turned out that planets fly along these curves, that telescopes collect light and dishes collect radio signals with them, and that an interstellar visitor that dropped into the Solar System in 2017 traced one of them. Let’s spend a night at the observatory and see why it’s these curves.

A torch on the wall

Let’s start with what we’re cutting. Take a line, the axis, and a second line crossing it at an acute angle $\alpha$. Rotate the second line about the axis. It sweeps out a surface of two identical funnels joined at their tips. That’s the cone, only not the kind that stands on a table, but an infinite double one. The lines it’s woven from are called generators, and the angle $\alpha$ is the half-angle of the cone.

What you get depends on the tilt of the plane. A plane perpendicular to the axis cuts the cone in a circle. Tilted slightly, in a closed oval, an ellipse. If the plane is parallel to one of the generators, that generator no longer meets it, and the curve runs off to infinity in a single branch: that’s a parabola. If the plane is tilted more steeply still, it cuts both funnels, and the curve falls into the two branches of a hyperbola. A separate case is a plane through the apex: then instead of a curve you get a point, a single line or a pair of lines. Such sections are called degenerate, and they’re mentioned only so as not to be surprised by them.

The earliest known use of these curves is linked to a problem from the chapter on the Elements: doubling the cube. Around 350 BC Menaechmus, a pupil of Eudoxus, noticed that if the curves $x^2 = ay$ and $y^2 = 2ax$ meet at a point $(x, y)$, then $x^4 = a^2y^2 = 2a^3x$, that is, $x^3 = 2a^3$. The edge of a cube of twice the volume is found, though not with compass and straightedge but by intersecting two conic sections. Before Apollonius these curves were obtained like this: three different cones were taken, acute-angled, right-angled and obtuse-angled, and each was cut by a plane perpendicular to a generator.

The gardener’s rope

A section of a cone is an honest definition but an awkward one: to draw an ellipse you don’t want to build a cone. Gardeners know a simpler way. Two pegs are driven into the ground and the ends of a rope, longer than the distance between the pegs, are tied to them. With a pointed stick the rope is pulled taut and taken round the pegs. The stick draws an oval flower bed on the ground.

The gardener’s pegs are the foci $F_1$ and $F_2$, and the sum of the distances $MF_1 + MF_2$ is the length of the rope. It’s usually written $2a$, and the distance between the foci $2c$. If the pegs are driven in at the same point, $c = 0$ and we get a circle of radius $a$: a circle is a special case of an ellipse.

To calculate, put the ellipse on the coordinate plane: the foci on the $x$-axis, symmetric about the origin, $F_1(-c, 0)$ and $F_2(c, 0)$. The condition “the sum of the distances is $2a$” can be written with the distance between points, but it gives an equation with two square roots, and squaring it twice is no great pleasure. There’s a shorter way.

The trick is to find first not the distances to the foci themselves but their difference: it can be expressed in terms of $x$ without a single root.

Write $r_1 = MF_1$ and $r_2 = MF_2$. The point $M$ lies on the ellipse when $r_1 + r_2 = 2a$. Drop from $M$ the perpendicular $MH$ to the $x$-axis; its foot is the point $H(x, 0)$. We get two right triangles $F_1HM$ and $F_2HM$ with the common leg $MH = |y|$. If $y = 0$ or $H$ falls on a focus, a triangle flattens into a segment, but the equations below still hold. The other legs are $F_1H = |x + c|$ and $F_2H = |x - c|$. By Pythagoras’ theorem $r_1^2 = (x + c)^2 + y^2$ and $r_2^2 = (x - c)^2 + y^2$. Subtract one from the other: the common leg drops out, and $r_1^2 - r_2^2 = (x + c)^2 - (x - c)^2 = 4cx$. A difference of squares factors: $r_1^2 - r_2^2 = (r_1 + r_2)(r_1 - r_2) = 2a\,(r_1 - r_2)$. So $r_1 - r_2 = \frac{2cx}{a}$. Knowing the sum and difference of two numbers, we find the numbers themselves: $r_1 = a + \frac{c}{a}x$ and $r_2 = a - \frac{c}{a}x$. The distances to the foci have turned out to be linear functions of $x$. Along the way, let’s find the meaning of the number $b$. At $x = 0$ we get $r_1 = r_2 = a$: the point of the ellipse on the $y$-axis is at distance $a$ from each focus. The triangle with vertices $O$, $F_2$ and this point is right-angled, with hypotenuse $a$ and leg $c$, so the other leg is $\sqrt{a^2 - c^2} = b$. The ellipse crosses the $y$-axis at the points $(0, \pm b)$. Square the equation $r_2 = a - \frac{c}{a}x$ and substitute $r_2^2 = (x - c)^2 + y^2$: $a^2 - 2cx + \frac{c^2}{a^2}x^2 = x^2 - 2cx + c^2 + y^2$. The $-2cx$ terms cancel, leaving $a^2 - c^2 = x^2\left(1 - \frac{c^2}{a^2}\right) + y^2$, that is, $b^2 = \frac{b^2}{a^2}x^2 + y^2$. Divide by $b^2$ and we get $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Conversely, let $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Then $|x| \le a$ and $y^2 = b^2 - \frac{b^2}{a^2}x^2$. Substitute this into $(x - c)^2 + y^2$ and go through the last step’s calculation backwards: we get $\left(a - \frac{c}{a}x\right)^2$. The number $a - \frac{c}{a}x$ is positive, because $\frac{c}{a}|x| \le c < a$, so $r_2 = a - \frac{c}{a}x$. Likewise $r_1 = a + \frac{c}{a}x$, and $r_1 + r_2 = 2a$: the point lies on the ellipse.

The semi-major axis: half the length of the rope, and the distance from the centre to the farthest points of the ellipse, the vertices $(\pm a, 0)$. The semi-minor axis: the ellipse crosses the $y$-axis at the points $(0, \pm b)$. From these points the distance to each focus is exactly $a$. Half the distance between the foci. At $c = 0$ the foci merge, $b = a$, and the equation becomes the equation of a circle, $x^2 + y^2 = a^2$. Example: the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$ has semi-axes $a = 5$ and $b = 3$, so $c = \sqrt{25 - 9} = 4$, and the foci are $(\pm 4, 0)$. Check at the point $(0, 3)$: the distance to each focus is $\sqrt{16 + 9} = 5$, together $10 = 2a$.

An ellipse is a squashed circle, and the equation says so outright. If the point $(x, y)$ lies on the circle $x^2 + y^2 = a^2$, then the point $\left(x, \frac{b}{a}y\right)$ lies on the ellipse: substitute and check. So the ellipse is obtained from the circle by squashing it towards the $x$-axis by a factor of $\frac{a}{b}$. That’s why a round saucer casts an elliptical shadow on the floor on a sunny day.

A gardener drove in the pegs $6$ m apart and took a rope $10$ m long. How wide will the flower bed be at its narrowest? Give the answer in metres.

Here $2a = 10$ and $2c = 6$, so $a = 5$, $c = 3$ and $b = \sqrt{25 - 9} = 4$. The narrowest place is the minor axis, of length $2b = 8$ m. If you got $4$, you found the semi-axis, that is, half the width.

A ruler, a set square and a string

Replace the sum in the definition of an ellipse with a difference.

The difference is written $2a$, the distance between the foci again $2c$, but now $a < c$. The points closer to $F_2$ form one branch, those closer to $F_1$ the other. A hyperbola is drawn with a ruler and a string. One end of the ruler is pinned to the focus $F_1$ so that it can turn. A string $2a$ shorter than the ruler is tied to the other end of the ruler, and the string’s other end to the focus $F_2$. A pencil presses the string against the ruler. The stretch of ruler from $F_1$ to the pencil is $2a$ longer than the path along the string from the pencil to $F_2$: that’s exactly the definition. The second tab of the widget above has such a ruler.

The calculation for the hyperbola repeats the one for the ellipse word for word, only now $r_1 - r_2 = \pm 2a$, so $r_1 + r_2 = \pm\frac{2cx}{a}$ and $r_2 = \pm\left(\frac{c}{a}x - a\right)$. Squaring gives the same equation $a^2 - c^2 = x^2\left(1 - \frac{c^2}{a^2}\right) + y^2$, but now the number $a^2 - c^2$ is negative, and it’s natural to write $b^2 = c^2 - a^2$.

The transverse semi-axis: the vertices of the hyperbola are the points $(\pm a, 0)$, and the difference of the distances to the foci is $2a$. The conjugate semi-axis: the hyperbola has no points on the $y$-axis, but $b$ sets the slope of the asymptotes, the lines $y = \pm\frac{b}{a}x$ that the branches hug ever more tightly as they run off to infinity. Half the distance between the foci $(\pm c, 0)$. For a hyperbola $c$ is greater than $a$. Example: $\frac{x^2}{9} - \frac{y^2}{16} = 1$, here $a = 3$, $b = 4$, $c = \sqrt{9 + 16} = 5$, and the asymptotes are $y = \pm\frac43 x$. The point $\left(5, \frac{16}{3}\right)$ lies on the hyperbola: its distance to the focus $(5, 0)$ is $\frac{16}{3}$, to the focus $(-5, 0)$ it’s $\sqrt{100 + \frac{256}{9}} = \frac{34}{3}$, and the difference is $\frac{18}{3} = 6 = 2a$.

The asymptotes can be seen from the equation. Rewrite it as $y^2 = \frac{b^2}{a^2}x^2 - b^2$: for large $x$ the subtracted $b^2$ is negligible compared with the rest, and $y$ is almost $\pm\frac{b}{a}x$. How “almost” becomes a precise statement is explained in the next chapter.

A parabola has one focus, and instead of the second focus a line appears.

The instrument for a parabola is a set square and a string. A rail is laid along the directrix, and the set square slides along it. A string as long as the square’s vertical side is tied to its top corner, and the string’s other end to the focus. A pencil presses the string against that side. The string from the corner to the pencil and on to the focus has the length of the side. The piece of the side from the corner to the pencil plus the piece from the pencil to the rail also has the length of the side. The first pieces are shared, so the distance from the pencil to the focus equals the distance to the rail. That’s the third tab of the widget.

Put the focus at the point $F\left(\frac p2, 0\right)$ and the directrix vertical, $x = -\frac p2$; the number $p > 0$ is the distance from the focus to the directrix. For a point $M(x, y)$ the distance to the directrix is $\left|x + \frac p2\right|$, and to the focus $\sqrt{\left(x - \frac p2\right)^2 + y^2}$. Equate the squares: $x^2 - px + \frac{p^2}{4} + y^2 = x^2 + px + \frac{p^2}{4}$, so $y^2 = 2px$. Conversely, $y^2 = 2px$ gives the same equality of squares, and the distances themselves are equal because both are non-negative.

The parameter: the distance from the focus to the directrix. The larger it is, the wider the parabola opens. The directrix. The vertex of the parabola, the origin, lies exactly halfway between the focus and the directrix. Example: $y^2 = 8x$, here $2p = 8$, $p = 4$. The focus is $(2, 0)$, the directrix $x = -2$. The point $(2, 4)$ lies on the parabola: $16 = 8 \cdot 2$. Its distance to the focus is $4$, to the directrix $2 + 2 = 4$.

And the parabola from the chapter on quadratic equations, the graph $y = x^2$, is the same thing, only turned: swap $x$ and $y$ and you get $x^2 = 2py$ with $2p = 1$. So the school parabola $y = x^2$ has the focus $\left(0, \frac14\right)$ and the directrix $y = -\frac14$. Check at the point $(1, 1)$: to the focus $\sqrt{1 + \frac{9}{16}} = \frac54$, to the directrix $1 + \frac14 = \frac54$. Any parabola $y = ax^2 + bx + c$ has a focus too: it’s obtained from $y = ax^2$ by a shift, and the focus travels with it.

Find the focus of the parabola $y = \frac{x^2}{8}$. Give the answer as the coordinates of a point.

Rewrite the equation: $x^2 = 8y = 2py$ with $p = 4$. The focus lies on the axis of symmetry, $\frac p2 = 2$ above the vertex: it’s the point $(0, 2)$, and the directrix is the line $y = -2$. Check at the point $(4, 2)$ of the parabola: the distance to the focus is $4$, to the directrix $2 + 2 = 4$.

A branch of a hyperbola and a parabola look very alike. Can a parabola be chosen to coincide with a branch of a hyperbola?

No: far out, a branch of a hyperbola is almost straight, while a parabola bends ever more steeply Yes, if it’s suitably shifted and rotated Yes, if you take a parabola with the right parameter $p$ It’s the same curve, just cut at different places on the cone

Running off to infinity, a branch of a hyperbola hugs the asymptote $y = \frac{b}{a}x$: the distance to that line gets arbitrarily small. A parabola has no such line. Take $y = x^2$ and any line $y = kx + m$: the difference $x^2 - kx - m$ grows without limit for large $x$, so the parabola pulls away from every line. A parabola has no straight tail; a hyperbola does.

Spheres in a cone

We now have two definitions of an ellipse, and they have nothing to do with each other. The first is a section of a cone: there are no foci in it. The second is the gardener’s rope: there’s no cone in it. Why is it the same curve? Apollonius knew the answer but reached it by long calculations. In 1822 the Belgian mathematician Germinal Dandelin found an argument that fits on one drawing.

The trick is to replace the distances to the foci with pieces of a generator: all the tangents from one point to a sphere are equal.

The plane cuts half of the cone in a closed curve. Take any point $M$ on it. We need two points whose distances to $M$ have the same sum for every point of the curve. Inscribe a sphere in the cone above the plane and inflate it until it touches the plane. The cone and the sphere are symmetric about the axis, so the sphere touches the cone along a circle $k_1$, and the plane at a single point $F_1$. On the other side of the plane inscribe a second sphere, also touching both the cone and the plane: the circle of contact is $k_2$, the point of contact $F_2$. Tangent segments drawn to a sphere from one point are equal. Indeed, if $MT$ touches a sphere with centre $O$ and radius $r$ at the point $T$, the radius $OT$ is perpendicular to the tangent, and by Pythagoras’ theorem $MT^2 = MO^2 - r^2$: this number doesn’t depend on the choice of the point of contact. The segment $MF_1$ lies in the plane that touches the first sphere at $F_1$, so it’s a tangent segment. The generator through $M$ touches the first sphere at a point $P$ of the circle $k_1$, so $MP$ is a tangent segment too. Hence $MF_1 = MP$. In the same way for the second sphere: $MF_2 = MQ$, where $Q$ is the point where the same generator touches the circle $k_2$. The cutting plane separates the spheres, so $M$ lies on the generator between $P$ and $Q$, and $MF_1 + MF_2 = MP + MQ = PQ$. But $PQ$ is the piece of the generator between the planes of the circles $k_1$ and $k_2$, and a rotation about the axis takes any generator to any other together with these circles. So $PQ$ is the same for all generators, and the sum $MF_1 + MF_2$ is constant: the section is an ellipse with foci $F_1$, $F_2$ and major axis $2a = PQ$.

For a hyperbola the same argument works: the spheres sit in different halves of the cone, $M$ no longer lies between $P$ and $Q$, and what’s constant is the difference $|MF_1 - MF_2| = PQ$. A parabola has only one sphere. But each sphere has one more find to offer: the plane containing its circle of contact cuts the plane of the section in a line. That’s the directrix. In the cone widget it’s drawn dashed on the flat view of the section.

Which section you get at a given tilt

Let $\beta$ be the angle between the cutting plane and the axis of the cone, and $\alpha$ the half-angle. Take a point $M$ of the section and the sphere that touches the plane at the focus $F$ and the cone along the circle $k$. Let $\pi$ be the plane of this circle, and $d$ the line where $\pi$ meets the cutting plane. As in the proof, $MF = MP$, where $P$ is the point of the generator on $k$. The distance $h$ from $M$ to the plane $\pi$ can be found two ways. Along the generator: it’s tilted at the angle $\alpha$ to the axis, so $h = MP\cos\alpha$. Within the cutting plane: the shortest path from $M$ to the line $d$ in this plane is the perpendicular $MD$, and it’s tilted at the angle $\beta$ to the axis, so $h = MD\cos\beta$. Hence $\frac{MF}{MD} = \frac{MP}{MD} = \frac{\cos\beta}{\cos\alpha}$, the same number for all points of the section. That’s the eccentricity: $e = \frac{\cos\beta}{\cos\alpha}$. A plane across the axis ($\beta = 90^\circ$) gives $e = 0$, one parallel to a generator ($\beta = \alpha$) $e = 1$, one parallel to the axis ($\beta = 0$) $e = \frac{1}{\cos\alpha} > 1$. In the cone widget $\alpha = 30^\circ$ and $\beta = 90^\circ - \varphi$, so $e = \frac{\sin\varphi}{\cos 30^\circ}$.

One number for all the curves

It’s not only the parabola that has a directrix. Go back to the proof of the standard equation. It turned out in passing that the distances from a point of the ellipse to the foci are linear functions: $r_2 = a - \frac{c}{a}x$. The number $\frac{c}{a}$ deserves a name of its own.

From the proof of the standard equation: $MF_2 = r_2 = a - \frac{c}{a}x = a - ex$. Take $e$ out of the bracket: $a - ex = e\left(\frac{a}{e} - x\right)$. Draw the vertical line $x = \frac{a}{e}$. Since $e < 1$, it’s to the right of the vertex $(a, 0)$ and doesn’t meet the ellipse. The distance from $M(x, y)$ to this line is the length of the horizontal segment $MD$. The point $M$ is to the left of the line, since $x \le a < \frac{a}{e}$, so $MD = \frac{a}{e} - x$. We’ve got $MF_2 = e \cdot MD$ for every point of the ellipse. For the other focus it’s the same: $MF_1 = a + ex = e\left(x + \frac{a}{e}\right)$, and $x + \frac{a}{e}$ is the distance to the line $x = -\frac{a}{e}$.

The converse is true too. Write the condition $MF_2 = e \cdot MD$ in coordinates: $(x - c)^2 + y^2 = (a - ex)^2$. Expand the brackets, remembering that $ae = c$: $x^2 - 2cx + c^2 + y^2 = a^2 - 2cx + e^2x^2$, so $x^2(1 - e^2) + y^2 = a^2 - c^2 = b^2$. Since $1 - e^2 = \frac{b^2}{a^2}$, this is the standard equation. For a hyperbola the calculation is the same, only $e > 1$ and the line $x = \frac{a}{e}$ lies between the centre and the vertex. And for a parabola it’s simply the definition with $e = 1$.

The circle stands apart in this rule: the closer $e$ is to zero, the further away the directrix goes.

Now Apollonius’ names can be decoded. Put the origin at the vertex nearest the focus, and point the $x$-axis towards the focus. Let $p$ be half the chord through the focus perpendicular to the axis; in the widget it’s the distance from the focus to the family’s common points. Then all three curves are written with one equation.

The area of a rectangle with sides $x$ and $2p$. The Greeks thought in areas, and compared the square on $y$ with this rectangle. The correction. When $e < 1$ it’s positive, and the square on $y$ is smaller than the rectangle: in Greek, a “falling short”, ellipsis. When $e = 1$ there’s no correction, the square equals the rectangle: an “application”, parabole. When $e > 1$ the correction is negative, the square is larger: an “excess”, hyperbole. Example: the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$ with $e = 0.8$ and $p = \frac{b^2}{a} = \frac95$. If $x$ is measured from the vertex $(-5, 0)$, the equation becomes $y^2 = 3.6x - 0.36x^2$. Check: at $x = 5$, the centre of the ellipse, $y^2 = 18 - 9 = 9$, and $y = \pm 3 = \pm b$.

Incidentally, about school curves. In the chapter on functions the graph $y = \frac1x$ was called a hyperbola. That’s not a metaphor. Rotate the coordinate axes by $45^\circ$, as in the chapter on symmetry: $x = \frac{X - Y}{\sqrt2}$, $y = \frac{X + Y}{\sqrt2}$. Then $xy = \frac{X^2 - Y^2}{2}$, and the equation $xy = 1$ becomes $\frac{X^2}{2} - \frac{Y^2}{2} = 1$: a hyperbola with $a = b = \sqrt2$ and $c = 2$. Its foci in the original coordinates are the points $(\sqrt2, \sqrt2)$ and $(-\sqrt2, -\sqrt2)$. Check at the point $(1, 1)$: the distances to the foci are $2 - \sqrt2$ and $2 + \sqrt2$, and the difference is $2\sqrt2 = 2a$.

Mirrors and dishes

The word “focus” is Latin for “hearth”. It was brought into geometry by Johannes Kepler in 1604, in a book on optics, five years before he published the discovery of elliptical orbits. The name is apt: if you make a mirror in the shape of a parabola and point it at the Sun, you can light a fire at the focus. A proof of this survives in Diocles’ treatise On Burning Mirrors, written around 200 BC. To understand why it happens, you need to know how light reflects from a curved mirror. A ray reflects just as it would from a flat mirror touching the curve at the point of incidence: the angle of incidence equals the angle of reflection, with angles measured from the tangent.

What remains is to understand what a tangent to an ellipse is. For a circle it’s a line that touches it at one point and doesn’t go inside. For an ellipse we’ll say the same. Points whose distances to the foci add up to less than $2a$ lie inside, more than $2a$ outside; a line through a point $M$ of the ellipse is called a tangent if all its other points lie outside. For an arbitrary curve this won’t do, and the general definition of a tangent has to wait for the chapter on the derivative. For conic sections it’s enough.

Draw through $M$ the line with equal angles and prove that it’s the tangent. The trick, as in Heron’s shortest-path problem, is to reflect one of the foci.

Draw through $M$ the line $\ell$ that makes equal angles with $MF_1$ and $MF_2$ (they’re marked on the drawing). The angle between $\ell$ and $MF_2$ isn’t zero, so $F_2$ doesn’t lie on $\ell$. Extend $F_1M$ beyond the point $M$. The angle between this extension and $\ell$ is vertical to the angle between $\ell$ and $MF_1$, so it equals the angle between $\ell$ and $MF_2$. So the reflection in $\ell$ takes the ray $MF_2$ to the extension of the segment $F_1M$. The image of the focus, the point $F_2'$, lies on this extension, and $MF_2' = MF_2$: reflection preserves distances. Take any other point $X$ on $\ell$. It lies on the axis of reflection, so $XF_2 = XF_2'$. By the triangle inequality $XF_1 + XF_2' \ge F_1F_2'$, with equality only if $X$ lies on the segment $F_1F_2'$. But $F_1F_2' = F_1M + MF_2' = F_1M + MF_2 = 2a$, and the line $\ell$ meets the line $F_1F_2'$ only at the point $M$. So for $X \ne M$ we get $XF_1 + XF_2 = XF_1 + XF_2' > 2a$: the point $X$ lies outside the ellipse. All the points of $\ell$ except $M$ are outside: it’s the tangent. It makes equal angles with $MF_1$ and $MF_2$, so a ray coming from $F_1$ to $M$ leaves, by the law of reflection, along $MF_2$, towards the other focus.

The rays from one focus not only gather at the other but arrive there at the same moment: every path $F_1M + MF_2$ has length $2a$. This is what “whispering galleries” with elliptical ceilings rely on: stand at one focus, and a whisper from the other is heard as if spoken right beside you, while people in the middle can’t make out a thing. Lithotripters, which have broken up kidney stones without surgery since the early 1980s, work on the same principle. A spark discharge at one focus of an ellipsoidal reflector creates a shock wave, the reflector gathers it at the other focus, and the patient is positioned so that the stone is there.

Inside the parabola lie the points closer to the focus than to the directrix, outside those closer to the directrix. The tangent, as for the ellipse, is a line through the point $M$ all of whose other points lie outside.

By the definition of a parabola $MF = MH$: triangle $FMH$ is isosceles. Draw $\ell$, the bisector of angle $FMH$. In an isosceles triangle the bisector of the apex angle is also the altitude and the median (chapter on the triangle), that is, the perpendicular bisector of the base $FH$. So the reflection in $\ell$ swaps $F$ and $H$. Take another point $X$ on $\ell$. It lies on the perpendicular bisector of $FH$, so $XF = XH$. Drop from $X$ the perpendicular $XK$ to the directrix. The slanting segment $XH$ is longer than the perpendicular $XK$ unless $H$ and $K$ coincide. And they could coincide only if $X$ lay on the line $MH$; but $\ell$ meets that line only at $M$, since the angle between them is half the angle $FMH$ and isn’t zero. So $XF = XH > XK$: the point $X$ is farther from the focus than from the directrix, that is, outside. All the points of $\ell$ except $M$ are outside, so $\ell$ is the tangent. A ray travelling parallel to the axis falls on the point $M$ along the line $HM$. The reflection in $\ell$ takes the line $MH$ to the line $MF$, so the reflected ray passes through the focus.

The converse is true too: a lamp at the focus gives a strictly parallel beam after reflection. That’s how searchlights and old car headlights are made. And a satellite dish uses the theorem itself: the signal from a satellite hanging thirty-six thousand kilometres up arrives as almost parallel rays, and the dish gathers it into the receiver standing at the focus. On home dishes the receiver usually hangs off to the side, not in front of the centre. Such a dish is cut from a large parabolic mirror not from the middle but from the edge: the focus stays where it was, and the receiver doesn’t block the dish’s signal. And at the solar furnace in Odeillo, in the French Pyrenees, a parabolic mirror more than fifty metres high gathers sunlight into a spot where the temperature exceeds $3000\ ^\circ$C.

A spherical mirror, a piece of a circle in cross-section, gathers rays only approximately. Near the axis the circle almost coincides with a parabola, but further out the rays miss and build up that very bright curve with a point. That’s why the mirrors of large telescopes aren’t ground to a sphere. The Hubble Space Telescope is built to the Ritchey–Chrétien design: both its large and small mirrors are hyperbolic.

Why a telescope needs a hyperbola

The hyperbola has its own reflective property: the tangent at a point $M$ bisects the angle $F_1MF_2$. The proof is the same as for the ellipse, only the reflection is in the bisector $\ell$ of the angle $F_1MF_2$ itself. The image $F_2'$ of the focus $F_2$ lands on the ray $MF_1$, and $MF_2' = MF_2$. For a point $X \ne M$ on $\ell$, by the triangle inequality, $|XF_1 - XF_2| = |XF_1 - XF_2'| < F_1F_2' = |MF_1 - MF_2| = 2a$, strictly, because $X$ doesn’t lie on the line $F_1F_2'$. All the points of $\ell$ except $M$ turn out to be between the branches, where the difference of distances is less than $2a$, and $\ell$ is the tangent. It follows that a ray heading for the focus $F_2$ reflects from the branch and passes through $F_1$. In the Cassegrain telescope (1672) the large parabolic mirror gathers a star’s light towards its focus, but in the way stands a small convex hyperbolic mirror that shares a focus with the large one. It sends the beam back through a hole in the large mirror to its own second focus, where the eyepiece or camera sits. The tube comes out much shorter.

Orbits

Before Kepler astronomers were sure that heavenly bodies moved in circles, and if circles didn’t help, in circles mounted on circles. Kepler inherited the best observations of his time, Tycho Brahe’s records of many years, and for several years he tried to fit the orbit of Mars into a circular scheme. The best of his models disagreed with the observations by eight minutes of arc, about a quarter of the Moon’s apparent diameter. Kepler trusted Tycho’s accuracy, and later wrote that these eight minutes alone had pointed the way to reforming the whole of astronomy. In the New Astronomy (1609) he stated two laws: the planets move in ellipses with the Sun at one focus, and the segment from the Sun to a planet sweeps out equal areas in equal times. The third law came in 1619: the squares of the orbital periods are as the cubes of the semi-major axes. Mars is on average $1.524$ times farther from the Sun than the Earth, and its year is $1.881$ times longer: $1.524^3 \approx 3.54 \approx 1.881^2$.

Why was the ellipse noticed only with Mars? Because the planets’ orbits are almost circular. Mars has $e \approx 0.093$, and its minor semi-axis is only $0.44\%$ shorter than its major one: $\sqrt{1 - e^2} \approx 0.9956$. By eye such an orbit can’t be told from a circle. What gives it away is something else: the Sun sits not at the centre but at a focus, shifted by $9\%$ of the semi-major axis.

Kepler derived his laws from observations. Why they’re true was explained by Isaac Newton in the Mathematical Principles of Natural Philosophy (1687). The very first proposition of Book I is the law of areas, and the proof there is a drawing.

Newton replaces the smooth pull with a sequence of instantaneous kicks, and everything comes down to triangles with equal bases and heights.

Divide time into equal intervals. Suppose that during each the body flies straight at a steady speed, and at the end of the interval gets an instantaneous kick directed towards $S$. In the first interval the body went from $A$ to $B$. Without the kick, in the second interval the body would have gone the same distance along the same line, from $B$ to $c$, with $Bc = AB$. Triangles $SAB$ and $SBc$ have equal bases on one line and a common vertex $S$, and so a common height. Their areas are equal: the area of a triangle is half the base times the height. The kick at $B$ adds to the displacement a term directed towards $S$. Displacements add by the parallelogram rule, and the body arrives at the point $C$ for which the segment $cC$ is parallel to $BS$. Triangles $SBc$ and $SBC$ share the base $SB$, and their vertices $c$ and $C$ lie on a line parallel to $SB$, that is, at the same distance from it. The heights are equal, the areas are equal. So triangles $SAB$ and $SBC$ have equal areas, and so on at every next step: equal areas in equal times. Newton makes the intervals shorter and shorter, the broken line approaches the smooth orbit, and the equality of areas carries over to it. What “approaches” means, and why the property survives such a passage, is explained in the next chapter.

Neither the ellipse nor the law of universal gravitation took part in the proof: the law of areas holds for any force directed towards one point. That with a force falling off as the square of the distance the orbit is a conic section, Newton proved separately, and that needs calculus. It also turns out that the ellipse isn’t the only option. A body arriving from afar at high speed will pass the Sun along a hyperbola and fly away for ever. That’s how ʻOumuamua, the first visitor spotted from another star system, passed through in 2017: the eccentricity of its path is about $1.2$. And in 2019 Gennady Borisov, with a telescope he had built himself, spotted an interstellar comet with $e \approx 3.4$.

From the law of areas it’s easy to find how many times faster a planet moves at perihelion, its closest point to the Sun, than at aphelion, its farthest. At both these points the velocity is perpendicular to the radius, and in a short time $\Delta t$ the radius sweeps out an almost right triangle with legs $r$ and $v\,\Delta t$, of area about $\frac12 rv\,\Delta t$. The areas in equal times are equal, so $r_{\min}v_{\max} = r_{\max}v_{\min}$, and $r_{\min} = a - c$ and $r_{\max} = a + c$. Hence

$$\frac{v_{\max}}{v_{\min}} = \frac{r_{\max}}{r_{\min}} = \frac{a + c}{a - c} = \frac{1 + e}{1 - e}.$$

For the Earth $e \approx 0.0167$, and the ratio is about $1.034$: about $147.1$ million km from the Sun at perihelion and $152.1$ million km at aphelion. For Halley’s comet, $e \approx 0.967$, it’s almost sixty: near the Sun it races along at about $55$ km/s, and beyond the orbit of Neptune it crawls at less than a kilometre per second.

When is the Earth closest to the Sun?

In summer, in July: that’s why it’s hot At the beginning of January At the equinoxes The distance is always the same

The Earth passes perihelion in the first days of January and aphelion at the beginning of July. The seasons are caused by the tilt of the Earth’s axis, not by the distance: a $3\%$ difference hardly affects the weather. But by the law of areas the Earth moves faster in January, and the winter half of the year in the northern hemisphere is shorter than the summer half: from the autumn equinox to the spring equinox there are about $179$ days, and from spring to autumn about $186$.

A comet moves along an ellipse with eccentricity $0.6$. How many times faster is it near the Sun than at the farthest point of its orbit?

$\frac{1 + e}{1 - e} = \frac{1.6}{0.4} = 4$. The aphelion is the same number of times farther from the Sun than the perihelion: if $a = 5$, then $c = 3$, and the distances are $2$ and $8$.

The observing log

The night at the observatory is coming to an end. Let’s gather the finds into one table: for each curve, the standard equation, the foci and the eccentricity.

The trainer has three levels: ellipses, parabolas, and hyperbolas mixed with the question “which curve is this?”.

Where next

We found the tangent to the ellipse and the parabola by a trick: through reflection and the triangle inequality. But try drawing a tangent to the graph of $y = x^3 - x$ or to a sine wave. These curves have no foci, and the trick won’t work. It’s the same with motion. The law of areas says a planet is faster at perihelion, but what is its speed at the very moment it passes perihelion? Average speed is distance divided by time. Speed “at a moment” is the distance covered in zero time divided by zero time: $\frac00$. Newton and Leibniz learned to handle such fractions with infinitely small quantities, and for a century and a half they were reproached for it being logically illegitimate. How to make it legitimate is told in the chapter on limits.