8. Coordinate geometry¶

Euclid proves theorems with clever constructions: draw this line, notice that pair of similar triangles. Each proof needs a new idea. In the 1630s two Frenchmen found a way to turn geometry into algebra: give every point two numbers, and every line and curve becomes an equation. Then many problems that needed a flash of insight can be solved by calculation. This chapter sets up coordinates, finds the equations of lines and circles, proves classical theorems by algebra, derives the parabola, ellipse and hyperbola from one focus-and-directrix definition with their reflection properties, and then follows Kepler to the shape of the orbit of Mars. It ends with polar coordinates, the natural language of orbits.

In this chapter

You will learn:

how Descartes and Fermat joined algebra to geometry, and the distance, midpoint and section formulas
the equations of lines and circles, perpendicular slopes, and the distance from a point to a line
how to prove geometry theorems with coordinates, and how to choose the axes to make it easy
the conic sections from the focus-directrix definition: their equations, eccentricity and reflection properties
Kepler's laws, and the real orbit of Mars (eccentricity 0.0934) with its perihelion, aphelion and the extra sunlight it brings
polar coordinates, some famous polar curves, and the polar equation of a conic

You need: Chapter 4 (Pythagoras), Chapter 5 (similar triangles), Chapter 6 (circles and tangents) and Chapter 7 (sine and cosine), plus the algebra of Algebra, Chapter 3 (lines) and Algebra, Chapter 5 (completing the square). For a gentler version, see the Topics lessons Distance and midpoint, Slope and the equation of a line and Equations of circles.

Levels: 8.1 to 8.3 are intermediate; 8.4 is intermediate; 8.5 to 8.7 are intermediate to advanced.

8.1 Descartes and Fermat¶

In 1637 René Descartes published his Discourse on Method with three appendices showing the method at work. One of them, La Géométrie, opens by saying that any problem of geometry can be reduced to one about lengths of lines, which can then be handled by arithmetic. Pierre de Fermat had reached the same idea independently; his short treatise on plane and solid loci, written by the mid-1630s and passed around in manuscript, was printed only in 1679, after his death. Fermat showed that every first-degree equation in two unknowns describes a straight line and every second-degree one a conic section; Descartes attacked harder curves and problems that the Greeks had left open.

Neither used our picture of two perpendicular number lines covering the plane with a grid. That came gradually, in the work of their successors. The idea had ancestors too: Apollonius described conics through relations between lengths measured along fixed lines, and Nicole Oresme in the fourteenth century drew graphs of changing quantities. What was new was the systematic translation: a curve is an equation, and solving equations is finding points.

The set-up. Choose a point \(O\), the origin, and two perpendicular number lines through it, the \(x\)-axis and the \(y\)-axis, with the same unit. Every point \(P\) of the plane has a pair of coordinates \((x, y)\): walk \(x\) units along the \(x\)-axis and \(y\) units parallel to the \(y\)-axis. Every pair of real numbers gives exactly one point. A curve is described by an equation in \(x\) and \(y\): the curve is the set of points whose coordinates make the equation true. In space, three perpendicular axes give three coordinates \((x, y, z)\).

From now on there are two pictures of every object, and the art is to move between them: a line is a picture and the equation \(y = 2x + 1\); the point where two lines cross is a picture and the solution of two equations.

8.2 Distance and midpoint¶

Theorem (distance formula). The distance between \(A = (x_1, y_1)\) and \(B = (x_2, y_2)\) is

\[AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.\]

Proof. If \(A\) and \(B\) are not on a horizontal or vertical line, let \(C = (x_2, y_1)\). Then \(AC\) is horizontal with length \(|x_2 - x_1|\), \(CB\) is vertical with length \(|y_2 - y_1|\), and the angle at \(C\) is right. By Pythagoras, \(AB^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2\). If \(A\) and \(B\) are on a horizontal line the formula gives \(|x_2 - x_1|\), which is right, and the same for a vertical line. \(\blacksquare\)

In space the same argument used twice gives \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\).

Theorem (midpoint and section formulas). The midpoint of \(AB\) is

\[M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right),\]

and more generally the point \(P\) on segment \(AB\) with \(AP : PB = m : n\) is \(P = \left(\frac{n x_1 + m x_2}{m + n}, \frac{n y_1 + m y_2}{m + n}\right)\).

Proof. Drop perpendiculars from \(A\), \(P\), \(B\) to the \(x\)-axis. They are parallel, so by the intercept theorem (Chapter 5) their feet divide the \(x\)-axis segment from \(x_1\) to \(x_2\) in the same ratio \(m : n\). The point dividing \([x_1, x_2]\) in that ratio is \(x_1 + \frac{m}{m+n}(x_2 - x_1) = \frac{n x_1 + m x_2}{m + n}\). The same for \(y\). For the midpoint, \(m = n = 1\). \(\blacksquare\)

Worked example. \(A = (1, 2)\), \(B = (7, 10)\). Then \(AB = \sqrt{36 + 64} = 10\) and \(M = (4, 6)\). The point two thirds of the way from \(A\) to \(B\) (\(m : n = 2 : 1\)) is \(\left(\frac{1 + 14}{3}, \frac{2 + 20}{3}\right) = (5, 22/3)\).

Worked example (a right angle by numbers). Is the triangle with vertices \(A = (1, 1)\), \(B = (4, 5)\), \(C = (8, 2)\) a right triangle? \(AB^2 = 9 + 16 = 25\), \(BC^2 = 16 + 9 = 25\), \(AC^2 = 49 + 1 = 50\). Since \(25 + 25 = 50\), the converse of Pythagoras (Chapter 4) says the angle at \(B\) is right, and the triangle is also isosceles.

8.3 Lines and circles in coordinates¶

Lines¶

The slope of the line through \((x_1, y_1)\) and \((x_2, y_2)\), with \(x_1 \neq x_2\), is

\[m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\text{rise}}{\text{run}} .\]

It does not depend on which two points of the line you pick: slope triangles on the same line are similar (their sides are parallel), so their rise-to-run ratios agree. In the language of Chapter 7, \(m = \tan\theta\) where \(\theta\) is the angle the line makes with the \(x\)-axis.

A point \((x, y)\) is on the line through \((x_1, y_1)\) with slope \(m\) exactly when \((y - y_1)/(x - x_1) = m\) or it is the point itself, so the line has the point-slope equation \(y - y_1 = m(x - x_1)\), which rearranges to \(y = mx + b\). Vertical lines have no slope and the equation \(x = k\). Every line has a general equation \(ax + by = c\) with \(a\) and \(b\) not both zero, and every such equation is a line. The Algebra book, Chapter 3 covers all these forms and how to solve two such equations together.

Theorem (parallel and perpendicular). Two non-vertical lines with slopes \(m_1\) and \(m_2\) are parallel exactly when \(m_1 = m_2\), and perpendicular exactly when \(m_1 m_2 = -1\).

Proof. Parallel: lines with the same slope make the same angle with the \(x\)-axis (equal corresponding angles), and conversely. Perpendicular: a line of slope \(m\) has a slope triangle with run 1 and rise \(m\). Rotate it by \(90°\) counterclockwise about its corner: the run of 1 to the right becomes a rise of 1 upward, and the rise of \(m\) upward becomes a run of \(m\) to the left. The rotated line is perpendicular to the first, and its slope is \(1/(-m) = -1/m\). So the perpendicular slope is \(m_2 = -1/m_1\), that is \(m_1 m_2 = -1\). (If \(m_1 = 0\) the perpendicular is vertical.) \(\blacksquare\)

Worked example. The line through \((2, 3)\) perpendicular to \(y = \tfrac12 x + 4\) has slope \(-2\): \(y - 3 = -2(x - 2)\), or \(y = -2x + 7\). The two lines meet where \(\tfrac12 x + 4 = -2x + 7\), at \(x = 1.2\), \(y = 4.6\).

Theorem (distance from a point to a line). The distance from \(P_0 = (x_0, y_0)\) to the line \(ax + by + c = 0\) is

\[d = \frac{|a x_0 + b y_0 + c|}{\sqrt{a^2 + b^2}} .\]

Proof. The shortest path from \(P_0\) to the line is along the perpendicular. The line \(ax + by + c = 0\) has slope \(-a/b\), so the perpendicular has slope \(b/a\): it runs in the direction \((a, b)\) (this also covers the cases \(a = 0\) or \(b = 0\), where the line is horizontal or vertical). Its points are \((x_0 + ta, y_0 + tb)\). The foot \(F\) is the one on the line:

\[a(x_0 + ta) + b(y_0 + tb) + c = 0, \qquad t = -\frac{a x_0 + b y_0 + c}{a^2 + b^2} .\]

The distance \(P_0F\) is \(|t| \sqrt{a^2 + b^2}\), which is the formula. \(\blacksquare\)

Worked example. The distance from \((3, 4)\) to \(3x + 4y - 5 = 0\) is \(|9 + 16 - 5|/5 = 4\).

Circles¶

By the distance formula, a point \((x, y)\) is on the circle with centre \((a, b)\) and radius \(r\) exactly when

\[(x - a)^2 + (y - b)^2 = r^2 .\]

Expanded, this has the form \(x^2 + y^2 + Dx + Ey + F = 0\). Going back, complete the square (Algebra, Chapter 5).

Worked example. \(x^2 + y^2 - 6x + 4y - 12 = 0\). Group and complete squares: \((x^2 - 6x + 9) + (y^2 + 4y + 4) = 12 + 9 + 4\), so \((x - 3)^2 + (y + 2)^2 = 25\): centre \((3, -2)\), radius 5. (If the right side had come out negative there would be no points; if zero, a single point.)

Lines meeting circles. Substitute the line into the circle and you get a quadratic; its discriminant says whether the line is a secant (two points), a tangent (one) or misses (none). For \(x^2 + y^2 = 25\) and \(y = x + 1\): \(x^2 + (x + 1)^2 = 25\), \(2x^2 + 2x - 24 = 0\), \(x^2 + x - 12 = 0\), \(x = 3\) or \(x = -4\), giving \((3, 4)\) and \((-4, -3)\). Check: \(9 + 16 = 25\) and \(16 + 9 = 25\).

Tangent lines. The tangent to \(x^2 + y^2 = r^2\) at a point \((x_1, y_1)\) on it is \(x_1 x + y_1 y = r^2\). Proof: the tangent is perpendicular to the radius (Chapter 6), which runs in the direction \((x_1, y_1)\), so the tangent's equation has the form \(x_1 x + y_1 y = c\) (its slope \(-x_1/y_1\) times the radius slope \(y_1/x_1\) is \(-1\)), and it passes through \((x_1, y_1)\), so \(c = x_1^2 + y_1^2 = r^2\). \(\blacksquare\) For example the tangent to \(x^2 + y^2 = 25\) at \((3, 4)\) is \(3x + 4y = 25\); by the formula above its distance from the origin is \(25/5 = 5\), the radius, as it should be.

The circle through three points. Its centre is where the perpendicular bisectors meet (Chapter 6). For \((0, 0)\), \((4, 0)\) and \((0, 2)\): the bisector of the first two is \(x = 2\), of the first and third \(y = 1\), so the centre is \((2, 1)\) and the radius \(\sqrt{5}\). Check: \((4 - 2)^2 + (0 - 1)^2 = 5\). (Or notice the right angle at the origin: by Thales the hypotenuse from \((4, 0)\) to \((0, 2)\) is a diameter, with midpoint \((2, 1)\).)

8.4 Proving theorems with coordinates¶

To prove a theorem with coordinates, place the figure so that the algebra is as simple as possible: put a vertex at the origin, a side along an axis, a centre of symmetry at the origin. That costs nothing, because distances and angles do not depend on where the axes are; but the figure must stay general: you may choose the axes, not the shape.

Theorem. The diagonals of a parallelogram bisect each other.

Proof. Put one vertex at \(O = (0, 0)\) and one side along the \(x\)-axis: \(A = (a, 0)\). Let the other side from \(O\) end at \(C = (b, c)\). The fourth vertex is \(B = (a + b, c)\), since \(CB\) must be parallel and equal to \(OA\). The midpoint of diagonal \(OB\) is \(\left(\frac{a + b}{2}, \frac c2\right)\) and the midpoint of diagonal \(AC\) is \(\left(\frac{a + b}{2}, \frac c2\right)\). They are the same point. \(\blacksquare\)

Theorem (Thales again). If \(AB\) is a diameter of a circle and \(C\) is another point of it, then \(AC \perp BC\).

Proof. Take the circle \(x^2 + y^2 = r^2\), \(A = (-r, 0)\), \(B = (r, 0)\), \(C = (x, y)\) with \(y \neq 0\). The slopes of \(AC\) and \(BC\) multiply to

\[\frac{y}{x + r} \cdot \frac{y}{x - r} = \frac{y^2}{x^2 - r^2} = \frac{y^2}{-y^2} = -1 . \qquad \blacksquare\]

Theorem (Apollonius' median theorem). If \(M\) is the midpoint of \(BC\) in triangle \(ABC\), then \(AB^2 + AC^2 = 2AM^2 + 2BM^2\).

Proof. Put \(M\) at the origin and \(BC\) on the \(x\)-axis: \(B = (-m, 0)\), \(C = (m, 0)\), \(A = (p, q)\). Then \(AB^2 + AC^2 = (p + m)^2 + q^2 + (p - m)^2 + q^2 = 2p^2 + 2q^2 + 2m^2 = 2AM^2 + 2BM^2\). \(\blacksquare\)

The same computation, read for a parallelogram, says that the sum of the squares of the four sides equals the sum of the squares of the two diagonals (the parallelogram law).

Theorem (centroid). The three medians of a triangle meet at one point, two thirds of the way from each vertex to the midpoint of the opposite side. This point, the centroid, is \(G = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right)\).

Proof. The median from \(A = (x_1, y_1)\) goes to the midpoint \(M = \left(\frac{x_2 + x_3}{2}, \frac{y_2 + y_3}{2}\right)\) of \(BC\). The point dividing \(AM\) in the ratio \(2 : 1\) is, by the section formula,

\[\left(\frac{1 \cdot x_1 + 2 \cdot \frac{x_2 + x_3}{2}}{3}, \ldots\right) = \left(\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}\right).\]

This expression is symmetric in the three vertices, so the same point lies two thirds of the way along each of the other two medians as well. \(\blacksquare\)

Worked example. For \(A = (0, 0)\), \(B = (6, 0)\), \(C = (3, 9)\) the centroid is \((3, 3)\). Check: the midpoint of \(BC\) is \((4.5, 4.5)\), and two thirds of the way from \((0, 0)\) to it is \((3, 3)\).

How far can this go? Descartes believed his method could solve any problem in geometry, and in a sense he was right: every statement about points, lines and circles defined by equations becomes a statement about polynomials, and in the twentieth century Tarski proved that there is an algorithm to decide all such statements. In practice the algebra can explode. Good coordinates, and later vectors (Chapter 9) and complex numbers (Chapter 13), keep it in hand; and synthetic proofs often show why a theorem is true where a calculation only shows that it is.

8.5 Conic sections¶

Shine a torch at a wall. Held straight at the wall, the patch of light is a circle; tilt it and the circle stretches into an ellipse; tilt it further until the top edge of the beam is parallel to the wall and the ellipse opens into a parabola; tilt it more and it becomes one branch of a hyperbola. These four curves are the conic sections, the curves in which a plane cuts a cone. Menaechmus, in the fourth century BCE, is credited by later writers with first studying them, in connection with the problem of doubling the cube. Apollonius of Perga, around 200 BCE, wrote the great treatise on them, the Conics, and gave them their names.

Rather than cutting cones, we use a flat definition that gives all three at once.

Definition (focus and directrix). Fix a line \(\ell\) (the directrix), a point \(F\) not on it (the focus), and a number \(e > 0\) (the eccentricity). The set of points \(P\) with

\[PF = e \cdot \operatorname{dist}(P, \ell)\]

is called an ellipse if \(e < 1\), a parabola if \(e = 1\), and a hyperbola if \(e > 1\).

The figure shows three conics sharing one focus and one directrix. As \(e\) shrinks toward 0 the ellipse becomes rounder; a circle is the limiting case \(e = 0\) (with the directrix infinitely far away).

The parabola¶

Put the focus at \(F = (0, p)\) and the directrix at \(y = -p\), with \(p > 0\). For \(P = (x, y)\), the condition \(PF = \operatorname{dist}(P, \ell)\) reads \(\sqrt{x^2 + (y - p)^2} = |y + p|\). Square both sides:

\[x^2 + y^2 - 2py + p^2 = y^2 + 2py + p^2, \qquad x^2 = 4py .\]

So \(y = x^2/(4p)\): every parabola is the graph of a quadratic, after moving the axes. Conversely \(y = x^2\) has \(4p = 1\), focus \((0, \tfrac14)\) and directrix \(y = -\tfrac14\). Check with the point \((1, 1)\): its distance to the focus is \(\sqrt{1 + (3/4)^2} = 5/4\), and to the directrix \(1 + 1/4 = 5/4\).

Theorem (reflection property of the parabola). Every ray parallel to the axis of a parabola, reflected at the curve (angle of incidence equals angle of reflection, measured from the tangent), passes through the focus.

Proof. Let \(P\) be on the parabola, \(D\) the foot of the perpendicular from \(P\) to the directrix, so \(PD = PF\). Let \(t\) be the bisector of the angle \(FPD\). Because triangle \(FPD\) is isosceles, \(t\) is the perpendicular bisector of \(FD\).

\(t\) is the tangent. Take any other point \(Q\) on \(t\). Since \(Q\) is on the perpendicular bisector of \(FD\), \(QF = QD\). But \(QD\) is not perpendicular to the directrix (only \(PD\) is, among segments from \(t\) to \(D\)), so \(QD\) is longer than the distance from \(Q\) to the directrix. So \(QF > \operatorname{dist}(Q, \ell)\), and \(Q\) is not on the parabola. The line \(t\) meets the parabola only at \(P\), and it does not cross it (all its other points are on the side where \(QF > \operatorname{dist}(Q,\ell)\), the outside), so it is the tangent.

The reflection. A ray parallel to the axis arrives along the line \(DP\) extended beyond \(P\). The angle it makes with \(t\) equals the angle between \(PD\) and \(t\) (vertical angles), which equals the angle between \(t\) and \(PF\) (\(t\) is the bisector). So the reflected ray leaves along \(PF\), toward the focus. \(\blacksquare\)

That is why satellite dishes, radio telescopes and the dish antennas that carry signals between Earth and spacecraft at Mars are paraboloids (a parabola spun about its axis): a faint signal arriving parallel to the axis is gathered at the focus, where the receiver sits. Run backwards, a lamp at the focus of a car headlight sends out a parallel beam.

The ellipse¶

Definition (two foci). An ellipse is the set of points \(P\) whose distances to two fixed points \(F_1\), \(F_2\) (the foci) add up to a constant \(2a\), larger than \(F_1F_2\).

This is the gardener's construction: tie the ends of a string of length \(2a\) to two pegs and draw with a pencil that keeps the string taut.

Equation. Put the foci at \((\pm c, 0)\) with \(0 \le c < a\). The condition is

\[\sqrt{(x + c)^2 + y^2} + \sqrt{(x - c)^2 + y^2} = 2a .\]

Move the second root across and square: \((x + c)^2 + y^2 = 4a^2 - 4a\sqrt{(x - c)^2 + y^2} + (x - c)^2 + y^2\), which simplifies to

\[a\sqrt{(x - c)^2 + y^2} = a^2 - cx . \qquad (*)\]

Square again: \(a^2(x^2 - 2cx + c^2 + y^2) = a^4 - 2a^2cx + c^2x^2\), that is \((a^2 - c^2)x^2 + a^2y^2 = a^2(a^2 - c^2)\). Put \(b^2 = a^2 - c^2\) and divide by \(a^2b^2\):

\[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, \qquad b^2 = a^2 - c^2 .\]

(Each squaring step can be reversed for points of this curve, so nothing extra crept in.) The numbers \(a\) and \(b\) are the semi-major and semi-minor axes: the ellipse crosses the axes at \((\pm a, 0)\) and \((0, \pm b)\). The point \((0, b)\) is at distance \(a\) from each focus, which is where \(b^2 + c^2 = a^2\) comes from.

Eccentricity and directrix. Divide \((*)\) by \(a\) and set \(e = c/a\):

\[PF_2 = a - ex = e\left(\frac{a}{e} - x\right).\]

Since \(|x| \le a < a/e\), the bracket is the distance from \(P\) to the line \(x = a/e\). So the two-foci ellipse is the focus-directrix conic with focus \(F_2\), directrix \(x = a/e\) and eccentricity \(e = c/a < 1\). The two definitions agree.

Theorem (reflection property of the ellipse). A ray from one focus, reflected at the ellipse, passes through the other focus.

Proof. At a point \(P\) of the ellipse, let \(t\) be the line through \(P\) making equal angles with \(PF_1\) and \(PF_2\) (the external bisector of angle \(F_1PF_2\)). Reflect \(F_2\) in \(t\) to get \(F_2'\). Because the angles are equal, \(F_1\), \(P\) and \(F_2'\) lie on one straight line, so \(PF_1 + PF_2 = PF_1 + PF_2' = F_1F_2'\). For any other point \(Q\) of \(t\), \(QF_1 + QF_2 = QF_1 + QF_2' > F_1F_2' = 2a\) by the triangle inequality (Chapter 2), so \(Q\) is outside the ellipse. Hence \(t\) touches the ellipse only at \(P\): it is the tangent, and it makes equal angles with the two focal lines, which is the reflection law. \(\blacksquare\)

This is Heron's shortest-path argument, and it explains whispering galleries: in an elliptical room a whisper at one focus is heard at the other. Medicine uses the same fact: some machines for breaking up kidney stones make a shock wave at one focus of an ellipsoidal reflector, with the stone placed at the other.

The hyperbola¶

Definition. A hyperbola is the set of points whose distances to two foci \(F_1\), \(F_2\) differ by a constant \(2a\), smaller than \(F_1F_2 = 2c\).

The same algebra as for the ellipse (with \(c > a\) and \(b^2 = c^2 - a^2\)) gives

\[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, \qquad b^2 = c^2 - a^2,\]

and again the focus-directrix form with \(e = c/a > 1\) and directrices \(x = \pm a/e\). For large \(x\) and \(y\) the 1 on the right hardly matters, and the curve approaches the two asymptotes \(y = \pm\frac{b}{a}x\).

The tangent at a point of a hyperbola bisects the angle between the lines to the foci (the internal bisector this time), so a ray aimed at one focus is reflected toward the other. Telescope makers use a hyperbolic secondary mirror in the Cassegrain design for exactly this. And because a constant difference of distances is a constant difference of arrival times, a ship that measures the time difference between radio signals from two stations knows it is on a certain hyperbola; that was the principle of the LORAN navigation system.

The general second-degree equation¶

Every equation \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\) (with \(A\), \(B\), \(C\) not all zero) describes a conic, possibly a degenerate one (two lines, one line, a point, or nothing). If it is a genuine curve, it is

an ellipse (or circle) if \(B^2 - 4AC < 0\),
a parabola if \(B^2 - 4AC = 0\),
a hyperbola if \(B^2 - 4AC > 0\).

The proof rotates the axes to remove the \(xy\) term (rotations are in Chapter 10, and in linear algebra this is diagonalising a symmetric matrix); \(B^2 - 4AC\) does not change under rotation, and then completing squares finishes the job. For example \(xy = 1\) has \(B^2 - 4AC = 1 > 0\): it is a hyperbola, turned by \(45°\), with the axes as asymptotes.

Why are cone sections and focus curves the same? In 1822 Germinal Dandelin gave a beautiful argument. Put a ball inside the cone, touching the cone along a circle and touching the cutting plane at a point \(F\). From any point \(P\) of the section, the distance to \(F\) equals the distance along the cone to the circle of contact (two tangent segments from \(P\) to the same ball are equal, as in Chapter 6). With a second ball on the other side of the plane, the sum of the two distances is the distance between the two circles of contact along the cone, the same for every \(P\). So the section is an ellipse with the two touching points as foci. The full argument is in Hilbert and Cohn-Vossen's book in the Further reading.

8.6 Kepler's ellipses and the orbit of Mars¶

For two thousand years astronomers described planetary motion with circles: circles on circles, off-centre circles, circles around moving points. Johannes Kepler inherited the most accurate naked-eye observations ever made, Tycho Brahe's, and set himself to fit the motion of Mars. It took him years. His best model with circles still missed some of Tycho's positions by up to about 8 minutes of arc, more than Tycho's observations could be wrong by, and Kepler refused to ignore the difference. In Astronomia nova (1609) he announced the first two of his laws, and the third followed in Harmonices mundi (1619):

Each planet moves on an ellipse with the Sun at one focus.
The line from the Sun to the planet sweeps out equal areas in equal times: the planet moves faster near the Sun.
The square of the period is proportional to the cube of the semi-major axis: \(T^2 \propto a^3\). In years and astronomical units (AU, the mean Earth-Sun distance, about 149.6 million km), \(T^2 = a^3\).

Mars was the right planet to study. Its orbit is more eccentric than that of any planet then known except Mercury, which is hard to observe because it never strays far from the Sun.

The numbers. Mars's orbit has semi-major axis \(a \approx 1.5237\) AU (about 227.9 million km) and eccentricity \(e \approx 0.0934\).

Perihelion (closest to the Sun): \(a(1 - e) \approx 1.3814\) AU \(\approx 206.6\) million km.
Aphelion (farthest): \(a(1 + e) \approx 1.6660\) AU \(\approx 249.2\) million km.
Semi-minor axis: \(b = a\sqrt{1 - e^2} \approx 0.99563\,a\). The orbit's shape differs from a circle by less than half a per cent.
Sun's offset: the Sun sits \(c = ae \approx 0.1423\) AU \(\approx 21.3\) million km from the centre of the ellipse.

Drawn to scale, as in the figure, the ellipse looks like a circle; what gives it away is that the Sun is clearly not at its centre. That offset, not the flattening, is what Kepler detected.

Period. By the third law, \(T = a^{3/2} = 1.5237^{1.5} \approx 1.881\) years \(\approx 687\) days, the length of a Martian year.

Speed and sunlight. At perihelion and aphelion the velocity is perpendicular to the Sun-Mars line, so by the equal-areas law (the area swept in a short time is about \(\tfrac12 \cdot r \cdot v \cdot \Delta t\)) the product \(r v\) is the same at both points. So Mars moves faster at perihelion by the factor \((1 + e)/(1 - e) \approx 1.206\), about 21 per cent. Sunlight falls off as the square of the distance, so at perihelion Mars receives \(\left(\frac{1 + e}{1 - e}\right)^2 \approx 1.45\) times the sunlight it gets at aphelion: about 45 per cent more. That difference is large enough to shape the Martian seasons; the southern summer, which comes near perihelion, is shorter and hotter than the northern one. For Earth, with \(e \approx 0.0167\), the same ratio is only about 1.07.

Close approaches. Earth's orbit is nearly circular at about 149.6 million km. When Earth passes between the Sun and Mars (an opposition) while Mars is near perihelion, the distance between them is roughly \(206.6 - 150 \approx 57\) million km; in August 2003 it was about 55.8 million km, the closest in tens of thousands of years. Near aphelion, oppositions are about 100 million km apart. This is why missions to Mars leave in windows about every 26 months, and why some windows are better than others.

8.7 Polar coordinates¶

A radar screen, a lighthouse beam, and an astronomer describing an orbit all locate a point by how far and which direction, not by \(x\) and \(y\).

Definition. The polar coordinates of a point \(P\) are \((r, \theta)\), where \(r\) is the distance \(OP\) and \(\theta\) is the angle from the positive \(x\)-axis to the ray \(OP\). From Chapter 7,

\[x = r\cos\theta, \qquad y = r\sin\theta, \qquad r = \sqrt{x^2 + y^2}, \qquad \tan\theta = \frac yx .\]

Unlike Cartesian coordinates, polar coordinates are not unique: \((r, \theta)\) and \((r, \theta + 360°)\) are the same point, and the origin is \((0, \theta)\) for every \(\theta\). It is often convenient to allow negative \(r\), meaning the point at distance \(|r|\) in the opposite direction. To find \(\theta\) from \((x, y)\), take \(\tan^{-1}(y/x)\) and correct by \(180°\) when \(x < 0\) (programming languages provide atan2(y, x), which does this).

Worked example. The point \((-3, 3)\) has \(r = \sqrt{18} = 3\sqrt2\) and \(\theta = 135°\) (second quadrant). The point with polar coordinates \((4, 300°)\) is \((4\cos 300°, 4\sin 300°) = (2, -2\sqrt3)\).

Curves. Some curves are much simpler in polar form.

\(r = a\): the circle of radius \(a\) about the origin.
\(\theta = \alpha\): a line through the origin.
\(r = 2a\cos\theta\): a circle through the origin. Multiply by \(r\): \(r^2 = 2ar\cos\theta\), which is \(x^2 + y^2 = 2ax\), or \((x - a)^2 + y^2 = a^2\), the circle with centre \((a, 0)\) and radius \(a\).
\(r = a(1 + \cos\theta)\): a cardioid (heart shape), the path of a point on a circle rolling around another of the same size.
\(r = \cos 2\theta\), \(r = \cos 3\theta\): roses, with 4 and 3 petals.
\(r = a\theta\): the spiral of Archimedes, whose turns are equally spaced, like the groove of a record.

The polar equation of a conic¶

Put the focus at the origin and the directrix at \(x = k\), to the right (\(k > 0\)). For a point \(P = (r\cos\theta, r\sin\theta)\) on the side of the directrix containing the focus, \(PF = r\) and \(\operatorname{dist}(P, \ell) = k - r\cos\theta\). The focus-directrix condition \(r = e(k - r\cos\theta)\) solves to

\[r = \frac{\ell}{1 + e\cos\theta}, \qquad \ell = ek .\]

The number \(\ell\) (the semi-latus rectum) is the value of \(r\) at \(\theta = 90°\). For an ellipse, \(\theta = 0°\) gives the nearest point \(r = \ell/(1 + e)\) and \(\theta = 180°\) the farthest \(r = \ell/(1 - e)\); these add to \(2a\), which gives \(\ell = a(1 - e^2)\). (With the directrix on the right, the nearest point is in the direction \(\theta = 0°\), toward it.)

For Mars, \(\ell = 1.5237 \times (1 - 0.0934^2) \approx 1.5104\) AU, so

\[r(\theta) \approx \frac{1.5104}{1 + 0.0934\cos\theta} \text{ AU},\]

with \(\theta\) measured from the direction of perihelion (astronomers call \(\theta\) the true anomaly). Check: \(r(0°) \approx 1.5104/1.0934 \approx 1.3814\) and \(r(180°) \approx 1.5104/0.9066 \approx 1.6660\), the perihelion and aphelion above.

This is the form in which Newton, in the Principia (1687), derived Kepler's ellipses from the law of gravitation: an inverse-square force gives exactly the curves \(r = \ell/(1 + e\cos\theta)\), ellipses for bound orbits, parabolas and hyperbolas for objects fast enough to escape. Spacecraft on their way to Mars ride these curves: an ellipse touching Earth's orbit at perihelion and Mars's orbit at aphelion is the classic minimum-energy transfer route.

Area in polar coordinates. A thin sector of angle \(\Delta\theta\) and radius \(r\) has area about \(\tfrac12 r^2\Delta\theta\) (Chapter 7), so areas swept by a moving radius are sums of \(\tfrac12 r^2\,\Delta\theta\). Kepler's second law says this sum grows at a constant rate, \(\tfrac12 r^2 \frac{d\theta}{dt} = \text{constant}\). Making that precise is calculus.

Check yourself¶

1. Find the distance between \((-3, 5)\) and \((5, -1)\), and their midpoint.

\(\sqrt{8^2 + 6^2} = 10\); midpoint \((1, 2)\).

2. Find the equation of the line through \((2, 3)\) perpendicular to \(2x - 5y = 7\).

The given line is \(y = \tfrac25 x - \tfrac75\), slope \(\tfrac25\). The perpendicular slope is \(-\tfrac52\): \(y - 3 = -\tfrac52(x - 2)\), or \(5x + 2y = 16\). Check: \((2, 3)\) gives \(10 + 6 = 16\).

3. How far is the point \((1, 2)\) from the line \(3x - 4y + 15 = 0\)?

\(|3 - 8 + 15|/\sqrt{9 + 16} = 10/5 = 2\).

4. Find the centre and radius of \(x^2 + y^2 + 4x - 10y + 13 = 0\).

\((x + 2)^2 + (y - 5)^2 = -13 + 4 + 25 = 16\): centre \((-2, 5)\), radius 4.

5. Find the circle through \((1, 1)\), \((5, 1)\) and \((1, 3)\).

The angle at \((1, 1)\) is right (one side horizontal, one vertical), so by Thales the segment from \((5, 1)\) to \((1, 3)\) is a diameter: centre \((3, 2)\), radius \(\sqrt{4 + 1} = \sqrt5\). Equation \((x - 3)^2 + (y - 2)^2 = 5\). Check \((1, 1)\): \(4 + 1 = 5\).

6. Prove with coordinates that the diagonals of a rhombus are perpendicular.

Take \(O = (0, 0)\), \(A = (a, 0)\), \(C = (b, c)\) with \(OC = OA\), so \(b^2 + c^2 = a^2\), and \(B = (a + b, c)\). Diagonal \(OB\) runs in direction \((a + b, c)\) and diagonal \(AC\) in direction \((b - a, c)\). Neither diagonal is vertical, since \(b = -a\) or \(b = a\) would force \(c = 0\), a flattened rhombus. So the product of slopes is \(\frac{c}{a + b} \cdot \frac{c}{b - a} = \frac{c^2}{b^2 - a^2} = \frac{c^2}{-c^2} = -1\).

7. Find the focus and directrix of the parabola \(y = x^2/8\), and check one point.

\(x^2 = 8y = 4py\) with \(p = 2\): focus \((0, 2)\), directrix \(y = -2\). The point \((4, 2)\) is at distance \(\sqrt{16 + 0} = 4\) from the focus and \(2 + 2 = 4\) from the directrix.

8. For the ellipse \(x^2/25 + y^2/9 = 1\), find the foci, the eccentricity and the directrices. Check the two-foci property at \((0, 3)\).

\(a = 5\), \(b = 3\), \(c = \sqrt{25 - 9} = 4\): foci \((\pm 4, 0)\), \(e = 0.8\), directrices \(x = \pm 25/4\). From \((0, 3)\) each focus is \(\sqrt{16 + 9} = 5\) away, total \(10 = 2a\).

9. For the hyperbola \(x^2/16 - y^2/9 = 1\), find the foci, eccentricity and asymptotes.

\(c = \sqrt{16 + 9} = 5\): foci \((\pm 5, 0)\), \(e = 5/4\), asymptotes \(y = \pm\tfrac34 x\).

10. Earth's orbit has \(a \approx 149.6\) million km and \(e \approx 0.0167\). Find its perihelion and aphelion distances and the ratio of sunlight at the two. Compare with Mars.

Perihelion \(149.6 \times 0.9833 \approx 147.1\) million km, aphelion \(149.6 \times 1.0167 \approx 152.1\) million km. Sunlight ratio \((1.0167/0.9833)^2 \approx 1.07\), about 7 per cent, against about 45 per cent for Mars.

11. Convert \(r = 4\sin\theta\) to Cartesian form and describe the curve.

Multiply by \(r\): \(x^2 + y^2 = 4y\), so \(x^2 + (y - 2)^2 = 4\): the circle with centre \((0, 2)\) and radius 2, passing through the origin.

12. Prove the reflection property of \(y = x^2/(4p)\) by coordinates: show that the tangent at \(P = (x_0, y_0)\) is \(y = \frac{x_0}{2p}x - y_0\), that it meets the axis at \(G = (0, -y_0)\), and that \(FG = FP\) for the focus \(F = (0, p)\). Why does this finish the proof?

Tangent: on the parabola, \(\frac{x^2}{4p} - \left(\frac{x_0}{2p}x - y_0\right) = \frac{x^2 - 2x_0x + x_0^2}{4p} = \frac{(x - x_0)^2}{4p}\) (using \(y_0 = x_0^2/(4p)\)), which is zero only at \(x = x_0\) and positive elsewhere: the line touches at \(P\) and stays below the curve, so it is the tangent. Axis: at \(x = 0\) the line gives \(y = -y_0\). Distances: \(FG = p + y_0\) and \(FP^2 = x_0^2 + (y_0 - p)^2 = 4py_0 + y_0^2 - 2py_0 + p^2 = (y_0 + p)^2\). So triangle \(FGP\) is isosceles with \(\angle FGP = \angle FPG\). The vertical ray through \(P\) is parallel to the axis \(GF\), so the angle it makes with the tangent equals \(\angle FGP\) (corresponding angles), hence equals \(\angle FPG\): incoming and reflected rays make equal angles with the tangent, and the reflected ray is \(PF\).

What you learned¶

Coordinates turn points into pairs of numbers and curves into equations; Descartes and Fermat introduced the idea in the 1630s.
Distance comes from Pythagoras, the midpoint and section formulas from the intercept theorem; slopes of perpendicular lines multiply to \(-1\); the distance from a point to a line is \(|ax_0 + by_0 + c|/\sqrt{a^2 + b^2}\).
Circles are \((x - a)^2 + (y - b)^2 = r^2\); completing the square finds centre and radius, and the tangent at \((x_1, y_1)\) to \(x^2 + y^2 = r^2\) is \(x_1x + y_1y = r^2\).
With well-chosen axes, theorems such as the parallelogram diagonals, Thales, Apollonius' median theorem and the centroid fall to algebra.
The focus-directrix definition gives the ellipse (\(e < 1\)), parabola (\(e = 1\)) and hyperbola (\(e > 1\)), with the equations \(x^2/a^2 \pm y^2/b^2 = 1\) and \(x^2 = 4py\) and reflection properties used in dishes, mirrors and medicine.
Mars moves on an ellipse with \(a \approx 1.524\) AU and \(e \approx 0.0934\): from 206.6 to 249.2 million km from the Sun, with about 45 per cent more sunlight at perihelion and a year of 687 days.
Polar coordinates \((r, \theta)\) make circles, roses and spirals simple, and a conic with focus at the origin is \(r = \ell/(1 + e\cos\theta)\), the equation of every Kepler orbit.

Next: Chapter 9 adds arrows to coordinates: vectors, the dot and cross products, and geometry in three dimensions.