5.11: Loci and conic sections

This section offers a brief introduction to certain classically important loci in the plane. The word locus here refers to the set of all points satisfying some simple geometrical condition; and all the examples in this section are based on the notion of distance from a point and from a line.

Given a point O and a positive real r, the locus of points at distance r from O is precisely the circle of radius r with centre O. If r < 0, then the locus is empty; while if r = 0, the locus consists of the point O alone.

Given a line m and a positive real r, the locus of all points at distance r from the line m consists of a pair of parallel lines - one either side of the line m. Given a circle of radius r, and a positive real number d < r; the locus of points at distance d from the circle consists of two circles, each concentric with the given circle (one inside the given circle and one outside). If d > r, the locus consists of a single circle outside the given circle.

Given two points A and B, the locus of points which are equidistant from A and from B is precisely the perpendicular bisector of the line segment AB. And given two lines m, n the locus of points which are equidistant from m and from n takes different forms according as m and n are, or are not, parallel.

• If m and n are parallel, then the locus consists of a single line parallel to m and n and half way between them.
• If m and n meet at X (say), then the locus consists of the pair of perpendicular lines through X, that bisect the four angles at X.

Problem 220 Given a point F and a line m, choose m as the x-axis and the line through F perpendicular to m as the y-axis. Let F have coordinates (0, 2a).

(i) Find the equation that defines the locus of points which are equidistant from F and from m.

(ii) Does the equation suggest a more natural choice of axes - and hence a simpler equation for the locus?

The locus, or curve, in Problem 220 is called a parabola; the point F is called the focus of the parabola, and the line m is called the directrix. In general, the ratio

"the distance from X to F” : “the distance from X to m"

is called the eccentricity of the curve. Hence the parabola has eccentricity e = 1.

The parabola has many wonderful properties: for example, it is the path followed by a projectile under the force of gravity; if viewed as the surface of a mirror, a parabola reflects the sun's rays (or any parallel beam) to a single point - the focus F. Since the only variable in the construction of the parabola is the distance “2a” between the focus and the directrix, we can scale distances to see that any two different-looking parabolas must in fact be similar to one another - just as with any two circles. (It is hard not to infer from the graphs that y = 10x² is a “thin” parabola, and that $y=\left(\frac{1}{10}\right)x^2$ is a “fat” parabola. But the first can be rewritten in the form 10y=(10x)², and the second can be rewritten in the form $\left(\frac{y}{10}\right)={\left(\frac{x}{10}\right)}^2$ , so each is a re-scaled version of Y = X².)

So far we have considered loci defined by some pair of distances being equal, or in the ratio 1 : 1. More interesting things begin to happen when we consider conditions in which two distances are in a fixed ratio other than 1 : 1.

Problem 221

(a) Given two points A, B, with AB = 6. Find the locus of all points X such that AX : BX = 2:1.

(b) Given points A, B, with AB = 2b and a positive real number f. Find the locus of all points X such that AX : BX = f : 1.

Problem 222

(a) Given points A, B, with AB = 2c and a real number a > c. Find the locus of all points X such that AX + BX = 2a.

(b) Given a point F and a line m, find the locus of all points X such that the ratio

distance from X to the point F : distance from X to the line m

is a positive constant e < 1.

(c) Prove that parts (a) and (b) give different ways of specifying the same curve, or locus.

Problem 223

(a) Given points A, B, with AB = 2c, and a positive real number a. Find the locus of all points X such that |AX - BX| = 2a.

(b) Given a point F and a line m, find the locus of all points X such that the ratio

distance from X to the point F : distance from X to the line m

is a constant e > 1.

(c) Prove that parts (a) and (b) give different ways of specifying the same curve, or locus.

Problem 221 is sometimes presented in the form of a mild joke.

Two dragons are sleeping, one at A and one at B. Dragon A can run twice as fast as dragon B. A specimen of homo sapiens is positioned on the line segment AB, twice as far from A as from B, and cunningly decides to crawl quietly away, while maintaining the ratio of his distances from A and from B (so as to make it equally difficult for either dragon to catch him should they wake).

The locus that emerges generally comes as a surprise: if the man sticks to his imposed restriction, by moving so that his position X satisfies XA = 2 . XB, then he follows a circle and lands back where he started! The circle is called the circle of Apollonius, and the points A and B are sometimes referred to as its foci.

The locus in Problem 222 is an ellipse - with foci A (or F = (-ae, 0)) and B (= (ae, 0)), and with directrix m (the line $y=-\frac{a}{e}$ ; the line y = a is the second directrix of the ellipse). The “focus-focus” description in part (a) is symmetrical under reflection in both the line AB and the perpendicular bisector of AB. The “focus-directrix” description in (b) is clearly symmetrical in the line through F perpendicular to m; but it is a surprise to find that the equation

$$\frac{x^2}{a^2}+\frac{y^2}{a^2\left(1-e^2\right)}=1$$

is also symmetrical under reflection in the y-axis. If we set b² = a²(1 - e²), the equation takes the form

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$$

which crosses the x-axis when x = ±a, and crosses the y-axis when y = +b. In its standard form, we usually choose coordinates so that b < a: the line segment from (-a, 0) to (a, 0) is then called the major axis, and half of it (say from (0,0) to (a, 0)) - of length a - is called the semi-major axis; the line segment from (0, -b) to (0, b) is called the minor axis, and half of it (say from (0, 0) to (0, b)) - of length b - is called the semi-minor axis.

The form of the equation shows that an ellipse is obtained from a unit circle by stretching by a factor “a” in the x-direction, and by a factor “b” in the y-direction. This implies that the area of an ellipse is equal to πab (since each small s by s square that arises in the definition of the “area” of the unit circle gets stretched into an “as by bs rectangle”). However, the equation tells us nothing about the perimeter of an ellipse. Attempts to pin down the perimeter of an ellipse gave rise in the 18^th century to the subject of “elliptic integrals”.

Like parabolas, ellipses arise naturally in many important settings. For example, Kepler (1571-1630) discovered that the planetary orbits are not circular (as had previously been believed), but are ellipses - with the Sun at one focus (a conjecture which was later explained by Isaac Newton (1642-1727)). Moreover, the tangent to an ellipse at any point X is equally inclined to the two lines XA and XB, so that a beam emerging from one focus is reflected at every point of the ellipse so that all the reflected rays pass through the other focus.

The curve in Problem 223 is a hyperbola - with foci A (or F = (-ae, 0)) and B (= (ae, 0)), and with directrix m (the line $y=-\genfrac{}{}{0.1ex}{}{a}{e}$; the line $y=\genfrac{}{}{0.1ex}{}{a}{e}$ is the second directrix of the hyperbola). The “focus-focus” description in part (a) is symmetrical under reflection in both the line AB and the perpendicular bisector of AB. The “focus-directrix” description in (b) is clearly symmetrical in the line through F perpendicular to m; but it is a surprise to find that the equation

$$\frac{x^2}{a^2}-\frac{y^2}{a^2\left(e^2-1\right)}=1$$

is also symmetrical under reflection in the y-axis. Like parabolas and ellipses, hyperbolas arise naturally in many important settings - in mathematics and in the natural sciences.

All these loci were introduced and studied by the ancient Greeks without the benefit of coordinate geometry and equations. They were introduced as planar cross-sections of a cone - that is, as natural extensions of straight lines and circles (since the doubly infinite cone is the surface traced out when one rotates a line about an axis through a point on that line). The equivalence of the focus-directrix definition in Problems 220, 222, and 223 and cross sections of a cone follows from the next problem. All five constructions in Problem 224 work with the doubly-infinite cone, which we may represent as x² +y² = (rz)² - although this representation is not strictly needed for the derivations. The surface of the double cone extends to infinity in both directions, and is obtained by taking the line y = rz in the yz-plane (where r > 0 is constant), and rotating it about the z-axis. Images of this rotated line are called generators of the cone; and the point they all pass through (i.e. the origin) is called the apex of the cone.

Problem 224 (Dandelinʹs spheres: Dandelin (1794-1847))

(a) Describe the cross-sections obtained by cutting such a double cone by a horizontal plane (i.e. a plane perpendicular to the z-axis). What if the cutting plane is the xy-plane?

Figure 7: Conic sections.

(b) (i) Describe the cross-section obtained by cutting such a cone by a vertical plane through the origin, or apex.

(ii) What cross-section is obtained if the cutting plane passes through the apex, but is not vertical?

(c) Give a qualitative description of the curve obtained as a cross-section of the cone if we cut the cone by a plane which is parallel to a generator: e.g. the plane y - rz = c.

(i) What happens if c = 0?

(ii) Now assume the cutting plane is parallel to a generator, but does not pass through the apex of the cone - so we may assume that the plane cuts only the bottom half of the cone. Insert a small sphere inside the bottom half of the cone and above the cutting plane, and inflate the sphere as much as possible - until it touches the cone around a horizontal circle (the “contact circle with the cone”), and touches the plane at a single point F. Let the horizontal plane of the “contact circle with the cone” meet the cutting plane in the line m. Prove that each point of the cross-sectional curve is equidistant from the point F and from the line m - and so is a parabola.

(d) (i) Give a qualitative description of the curve obtained as a cross-section of the cone if we cut the cone by a plane which is less steep than a generator, but does not pass through the apex - and so cuts right across the cone.

(ii) We may assume that the plane cuts only the bottom half of the cone. Insert a small sphere inside the bottom half of the cone and above the cutting plane (i.e. on the same side of the cutting plane as the apex of the cone), and inflate the sphere as much as possible - until it touches the cone around a horizontal circle, and touches the plane at a single point F. Let the horizontal plane of the contact circle meet the cutting plane in the line m. Prove that, for each point X on the cross-sectional curve, the ratio

"distance from X to F” : “distance from X to m” = e : 1

is constant, with e < 1, and so is an ellipse.

Figure 8: The conic section arising in Problem 224(d).

(e) (i) Give a qualitative description of the curve obtained as a cross-section if we cut the cone by a plane which is steeper than a generator, but does not pass through the apex (and hence cuts both halves of the cone)?

(ii) We can be sure that the plane cuts the bottom half of the cone (as well as the top half). Insert a small sphere inside the bottom half of the cone and on the same side of the cutting plane as the apex, and inflate the sphere as much as possible - until it touches the cone around a horizontal circle, and touches the plane at a single point F. Let the horizontal plane of the contact circle meet the cutting plane in the line m. Prove that, for each point X on the cross-sectional curve, the ratio

“distance from X to F” : “distance from X to m” = e : 1

is constant, with e > 1, and so is a hyperbola.

Problem 224 reveals a remarkable correspondence. It is not hard to show algebraically that any quadratic equation in two variables x, y represents either a point, or a pair of crossing (possibly identical) straight lines, or a parabola, or an ellipse, or a hyperbola: that is, by changing coordinates, the quadratic equation can be transformed to one of the standard forms obtained in this section. Hence, the possible quadratic curves are precisely the same as the possible cross-sections of a cone. This remarkable equivalence is further reinforced by the many natural contexts in which these conic sections arise.