The Ellipse That Aims for You, math that surprises you lesson

Picture a billiard table shaped like an ellipse, with no pockets, and two special points inside it called the foci. Shoot a ball from one focus in any direction you like. After it bounces off the cushion, it travels straight through the other focus, whichever direction you chose. This page explains why by connecting the way an ellipse is drawn with a string to the rule that a bounce makes equal angles, and it states exactly which idealizations the claim needs.

Key terms

Reflection property of the ellipse

Definition Let an ellipse have foci $F_1$ and $F_2$ , so that every point $P$ on it has $PF_1 + PF_2 = 2a$ for one fixed length $2a$ , and fix such a point $P$ . The reflection property says that the segments $PF_1$ and $PF_2$ make equal angles with the tangent line at $P$ , the two angles lying on opposite sides of $P$ along the tangent. Equivalently, the line through $P$ perpendicular to the tangent splits the angle $F_1 P F_2$ into two equal halves. In billiard terms, a ball that leaves $F_1$ in any direction and bounces off the ellipse by the law of reflection passes through $F_2$ after the bounce, on the ideal table described in the elliptical billiards card. The reason is the definition of the ellipse. Extend the segment from $F_1$ through $P$ to the point $R$ beyond $P$ with $PR = PF_2$ , so that $F_1 R = PF_1 + PF_2 = 2a$ . Let $\ell$ be the perpendicular bisector of the segment $F_2 R$ , which passes through $P$ because $P$ is equally far from $F_2$ and $R$ . Every other point $Q$ of $\ell$ is also equally far from $F_2$ and $R$ , and it does not lie on the straight line through $F_1$ and $R$ , which meets $\ell$ only at $P$ . So

\begin{aligned} QF_1 + QF_2 &= QF_1 + QR \\[0.4em] &> F_1 R = 2a . \end{aligned}

Hence every point of $\ell$ except $P$ lies outside the ellipse, and a line that touches the ellipse at $P$ and stays outside it everywhere else is the tangent at $P$ . Reflection across $\ell$ swaps $R$ and $F_2$ , so it carries the straight continuation $PR$ of the incoming path onto the segment $PF_2$ . A mirror image makes the same angle with the mirror as the original, so the incoming path and $PF_2$ make equal angles with the tangent, which is exactly what the law of reflection asks for.

In plain words At every point of an ellipse, the tangent is tilted just right to act as a mirror between the two foci. The tilt comes from the string. Sliding away from $P$ along the tangent always makes the two distances to the foci add up to more than the length of the string.

Example Take the ellipse

\frac{x^2}{25} + \frac{y^2}{9} = 1

with foci $F_1 = (-4, 0)$ and $F_2 = (4, 0)$ , and the point $P = (4, 9/5)$ directly above $F_2$ . The tangent at a point $(x_0, y_0)$ of this ellipse is $x_0 x/25 + y_0 y/9 = 1$ , which at $P$ becomes $4x + 5y = 25$ . To check that this line touches the ellipse only at $P$ , substitute $y = (25 - 4x)/5$ into $9x^2 + 25y^2 = 225$ ,

\begin{aligned} 9x^2 + (25 - 4x)^2 &= 225 \\[0.4em] 25x^2 - 200x + 400 &= 0 \\[0.4em] 25(x - 4)^2 &= 0 , \end{aligned}

which has the single solution $x = 4$ . The tangent falls with slope $-4/5$ , so it makes an angle of about $38.66^\circ$ with the horizontal. The segment $PF_2$ is vertical, so it meets the tangent at $90^\circ - 38.66^\circ = 51.34^\circ$ . The segment from $F_1$ to $P$ rises with slope $9/40$ , about $12.68^\circ$ above the horizontal, and it meets the tangent at $12.68^\circ + 38.66^\circ = 51.34^\circ$ . The angles agree, so a ball from $F_1$ that hits the cushion at $P$ goes straight down through $F_2$ .

Foci of an ellipse

Definition Choose two points $F_1$ and $F_2$ in the plane, a distance $2c$ apart, and a length $2a$ with $a > c \ge 0$ . The ellipse with foci $F_1$ and $F_2$ is the set of all points $P$ with

PF_1 + PF_2 = 2a .

Points with a smaller sum lie inside the ellipse and points with a larger sum lie outside it. With the center at the origin and the foci at $(\pm c, 0)$ , the ellipse has the equation

\begin{gathered} \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \\[0.4em] b^2 = a^2 - c^2 . \end{gathered}

The long axis, the major axis, runs from $(-a, 0)$ to $(a, 0)$ and contains both foci, and the short axis, the minor axis, runs from $(0, -b)$ to $(0, b)$ . When $c = 0$ both foci sit at the center and the ellipse is a circle of radius $a$ . The condition $a > c$ matters, since a sum equal to $2c$ gives only the segment $F_1 F_2$ and a smaller sum gives no points at all.

In plain words Pin the two ends of a string of length $2a$ at the foci, pull it tight with a pencil and move the pencil around. The pencil draws the ellipse. As it moves, one piece of string gets longer and the other gets shorter by the same amount.

Example With $a = 5$ and $c = 4$ we get $b = 3$ , the ellipse $x^2/25 + y^2/9 = 1$ with foci $(\pm 4, 0)$ . At $(3, 12/5)$ the distance to $F_1 = (-4, 0)$ is $\sqrt{49 + 144/25} = 37/5$ and the distance to $F_2 = (4, 0)$ is $\sqrt{1 + 144/25} = 13/5$ . Other points give the same total.

Point	Distance to $F_1$	Distance to $F_2$	Sum
$(0, 3)$	$5$	$5$	$10$
$(3, 12/5)$	$7.4$	$2.6$	$10$
$(4, 9/5)$	$8.2$	$1.8$	$10$
$(5, 0)$	$9$	$1$	$10$

Law of reflection

Definition When an ideal ball, or a ray of light, hits a smooth wall at a point $P$ , it leaves along the path that makes the same angle with the tangent line of the wall at $P$ as the incoming path did, on the other side of the normal, the line through $P$ perpendicular to the tangent. Equivalently, the angle of incidence equals the angle of reflection, both measured from the normal, and the outgoing path is the mirror image, across the tangent line, of the straight continuation of the incoming path. The speed does not change. For a curved wall only the tangent line at the point of impact matters, so at that instant the wall acts like a flat mirror along its tangent. A path that meets the wall along the normal comes straight back, and running any reflected path backward gives another reflected path.

In plain words The wall turns the ball the way a mirror turns light, with the angle going in equal to the angle coming out. On a curved cushion, the mirror to use is the tangent line at the point of impact.

Example A ball travels from $(0, 3)$ to a straight wall along the $x$ -axis and hits it at $(4, 0)$ . Without the wall it would continue to $(8, -3)$ . The mirror image of that point across the wall is $(8, 3)$ , so the ball leaves toward $(8, 3)$ , and both paths make an angle of about $36.87^\circ$ with the wall, the angle whose tangent is $3/4$ . The route from $(0, 3)$ to $(4, 0)$ and on to $(8, 3)$ has length $5 + 5 = 10$ , the same as the straight segment from $(0, 3)$ to the mirror point $(8, -3)$ . Any other point of the wall gives a longer route, since its route has the same length as a bent path to $(8, -3)$ . The same mirror idea proves the reflection property of the ellipse.

Elliptical billiards

Definition Elliptical billiards is the mathematical model of a billiard table whose cushion is an exact ellipse. The ball is a point that moves in straight lines at constant speed, and at the cushion it reflects by the law of reflection with no loss of speed. There is no friction, spin, pocket, ball size or cushion that gives way. In this model the reflection property gives three facts.

A path through one focus passes through the other focus after the next bounce, then through the first focus again after the bounce after that, and so on.
Every path from $F_1$ to the cushion and on to $F_2$ has the same length $2a$ , so balls launched together from $F_1$ at the same speed reach $F_2$ together after one bounce.
A path that misses both foci on one stretch between bounces misses them on every stretch, because a single stretch through a focus would force all the earlier and later stretches through a focus as well.

Starting points other than the foci, and oval tables that are not ellipses, do not have this aiming property in general. On a real table the center of a ball of radius $r$ bounces off a curve lying $r$ inside the cushion, and that curve is not exactly an ellipse, so the exact argument no longer applies.

In plain words On a perfect elliptical table a shot from one focus is a sure shot through the other, however you aim it. The guarantee belongs to the two foci and to the exact shape of the cushion.

Example On the table $x^2/25 + y^2/9 = 1$ , a shot from $F_1 = (-4, 0)$ toward the top point $(0, 3)$ travels $5$ to the cushion and then, by symmetry, $5$ more down to $F_2 = (4, 0)$ . A shot toward $(4, 9/5)$ travels $8.2$ to the cushion and then $1.8$ straight down to $F_2$ . A shot along the axis to the right passes $F_2$ after $8$ , hits $(5, 0)$ after $9$ , comes straight back and passes $F_2$ again after $10$ . A shot to the left hits $(-5, 0)$ after $1$ and comes back to reach $F_2$ after $1 + 9 = 10$ . Every bounce path to $F_2$ has length $10 = 2a$ .

Check yourself

A table has the shape $x^2/169 + y^2/25 = 1$ . Where are its foci? Find the length of the path from one focus off the cushion at $(0, 5)$ to the other focus, and the same for the cushion point $(12, 25/13)$ .

Show the answer

Here $a = 13$ and $b = 5$ , so $c^2 = 169 - 25 = 144$ and the foci are $(\pm 12, 0)$ . The point $(0, 5)$ is at distance $\sqrt{144 + 25} = 13$ from each focus, so the path has length $26$ . The point $(12, 25/13)$ lies on the ellipse, since

\begin{aligned} &\frac{144}{169} + \frac{(25/13)^2}{25} \\[0.4em] &= \frac{144}{169} + \frac{25}{169} = 1 . \end{aligned}

It is directly above the focus $(12, 0)$ , at distance $25/13$ . Its distance to $(-12, 0)$ is

\sqrt{24^2 + \left(\frac{25}{13}\right)^2} = \frac{313}{13},

and $313/13 + 25/13 = 338/13 = 26$ . Both paths have length $2a = 26$ , as every one-bounce path between the foci does.

Change a rule and start the ball at the center $O$ of the table $x^2/25 + y^2/9 = 1$ instead of at a focus. In which directions does the ball pass through the center again right after its first bounce?

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Only in the four directions along the axes. After a bounce at $P$ the ball can pass through $O$ again only by traveling back along the line $PO$ , which means it reverses. The law of reflection reverses a path only when the path meets the cushion along the normal, the line perpendicular to the tangent. The tangent at $(x_0, y_0)$ is $x_0 x/25 + y_0 y/9 = 1$ , so the normal points along $(x_0/25, y_0/9)$ , and this is parallel to the direction $(x_0, y_0)$ of the segment from $O$ only when $x_0 = 0$ or $y_0 = 0$ . So the shot must hit one of the four points $(\pm 5, 0)$ and $(0, \pm 3)$ .

For example, a shot from $O$ toward $(4, 9/5)$ would continue to $(8, 18/5)$ , whose mirror image across the tangent $4x + 5y = 25$ is $(128/41, -512/205)$ . The ball leaves along the line toward that point, which crosses the major axis at $x = 3200/881 \approx 3.63$ , missing both the center and the focus $(4, 0)$ . The proof of the reflection property needs the two distances from $P$ to add up to the same total all around the ellipse, and the foci are the only pair of points with that property.

Extend the pattern. A ball leaves $F_1 = (-4, 0)$ on the table $x^2/25 + y^2/9 = 1$ and first hits the cushion at $(4, 9/5)$ . Find its next two bounce points and describe where the path goes after that.

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After the first bounce the ball goes straight down through $F_2 = (4, 0)$ and hits the cushion again at $(4, -9/5)$ . That stretch passed through $F_2$ , so by the reflection property the next stretch passes through $F_1$ . Its points are

\begin{gathered} x = 4 - 8t, \\[0.4em] y = -\tfrac95 + \tfrac95 t, \end{gathered}

with $t = 0$ at the bounce point and $t = 1$ at $F_1$ . Substituting into the equation of the ellipse and multiplying by $25$ gives $73t^2 - 82t = 0$ . The root $t = 0$ is the bounce just made, so the next bounce has $t = 82/73$ , at the point

\left( -\frac{364}{73}, \frac{81}{365} \right),

about $(-4.986, 0.222)$ . The path keeps alternating through the two foci, and the following bounces are at about $(4.9998, -0.0247)$ and $(-4.999998, 0.0027)$ . From the third bounce on, the bounce points get closer and closer to the two ends of the major axis, and the path approaches the straight back and forth motion along that axis. This happens for every path through a focus that does not already run along the major axis.