Chapter 18

Apollonius’ extremum problems on conics

Book V of Apollonius’ Conics treats minimum and maximum lines for conics, in particular the parabola. Let B(b, 0) be a point on the axis of the parabola y2 = 4ax. To ﬁnd a point P on the parabola such that BP is minimum. Let P be the point (at2, 2at) on the parabola. The square distance

BP 2 = (at2 b)2 + (2at)2 − = a2t4 2a(b a)t2 + b2 − − = (at2 (b a))2 + a(a 2b) − − − is a quadratic in t2.

P

2a

O F Q B

(1) If b a, this is minimum when t = 0, i.e., P is the vertex of the parabola, and≤ the minimum BP = b. 214 Apollonius’ extremum problems on conics

(2) If b>a, this is minimum when b a = at2, i.e., if BP is mini- mum and Q is the orthogonal projection of−P on the axis, then BQ = 2a. This is Proposition V.8 of Conics. In V.27, Apollonius shows that BP is perpendicular to the tangent at P . In other words, BP is a normal to the parabola. Apollonius also considers the same problem for ellipses and hyper- x2 y2 bolas. In the case of the ellipse a2 + b2 = 1, let B be the point (u, 0). For P (a cos t, b sin t) on the ellipse, BP 2 = (a cos t u)2 +(b sin t)2 − = ··· au 2 b2(c2 u2) = c2 cos t + − , − c2 c2 2 2 2 where c = a b . The foci F and F ′ of the ellipse are the points − c2 ( c, 0). Let Q and Q′ be the points with coordinates , 0 . ± ± a (1) If B is between Q and Q′, then the minimum point P (a cos t, b sin t) u has parameter t given by cos t = OQ . In this case, the line BP is per- pendicular to the tangent at P . (2) If B is outside the segment QQ′, then the minimum point is one of the vertices A, A′ of the ellipse.

P

′ t A A ′ ′ O Q F Q B F Chapter 19

Normals of a parabola

Given a point A with coordinates (x, y), what is the point on the parabola P : y2 = 4ax closest to A? If P is the point t on the parabola, the square distance of AP is given by F (t)= (at2 x)2 + (2at y)2. − − Consider the derivatives of F as a function of t:

2 F ′(t) = 2at(at x) + 2a(2at y) − − = 2a(at3 + 2at tx y), 2 − − F ′′(t) = 2a(3at + 2a x). − If F (t) is minimum, then F ′(t) = 0. This condition is exactly the same as the equation of the normal at t. Thus, if AP is minimum, then P lies on the normal through A. In general, through a point (x, y), there are 3 normal lines. The cubic equation at3 + (2a x)t y = 0 − − has a double root if and only if

2a x 3 y 2 − + = 0, 3a 2a or 27ay2 = 4(x 2a)3. − This is a semicubical parabola Q parametrized by x = 2a + 3as2, y = 2as3. 216 Normals of a parabola

The point Q(s) lies on the normals of P at P ( s) and P (2s). Since − F ′′( s) = 0 and F ′′(2s) > 0, P ( s) is a point of inﬂection and P (2s) a minimum− for F (t). −

P (2s)

Q(s)

O F 2a

P (−s)

Q(−2s)

Proposition 19.1. The normal at P (t) to P 2 (i) intersects P again at P (t′), where t′ = t , − − t (ii) is tangent to Q at Q( t), and intersects it at Q t . − − 2 Exercise (1) Find the normal to y2 = 4ax which passes through P (t). (2) Find the minimum normal chords of the parabola y2 = 4ax. (3) Calculate the length of the semicubical parabola Q between the points Q(0) and Q(s). Chapter 20

Envelope of normal to a conic

What we have found in 19 for parabolas extend to parametric curves. Consider a curve C with§ parametrization

x = f(t), y = g(t).

At the point t on the curve, the tangents and the normals are the lines

g′(t)(x f(t)) f ′(t)(y g(t))= 0, − − − f ′(t)(x f(t)) + g′(t)(y g(t))= 0. − − We consider the problem of minimizing the distance of a point on the curve from a given point P (x, y). The square distance is given by

F (t)=(x f(t))2 +(y g(t))2. − −

dF (t) = 2(f ′(t)(x f(t)) + g′(t)(y g(t)). dt − − −

This means that t is a critical point of F if and only if the normal at t contains P .

The envelope of the normal Consider the normal of C at the point t, with equation given by

f ′(t)(x f(t)) + g′(t)(y g(t))=0. − − 218 Envelope of normal to a conic

At a neighboring point t′, the normal is given by the same equation with t replaced by t′. If we write t′ = t + ε for a small ε, and replace f(t′) by f(t)+ f ′(t)ε, f ′′(t) by f ′(t)+ f ′′(t)ε, and similarly for g(t) and g′(t), then the normal at this neighboring point is the line

f ′(t)(x f(t)) + g′(t)(y g(t)) − − + ε(f ′′(t)(x ′ f(t)) + g′′(t)(y g′(t)) − − = 0. Therefore, solving the system of equations

f ′(t)(x f(t)) + g′(t)(y g(t))= 0, − − f ′′(t)(x f ′(t)) + g′′(t)(y g′(t))= 0, − − we obtain the intersection of two neighoring normals, namely,

f ′(t)2+g′(t)2 x = f(t) g′(t) f ′(t)g′′(t) f ′′(t)g′(t) , − · f ′(t)2−+g′(t)2 (20.1) (y = g(t)+ f ′(t) f ′(t)g′′(t) f ′′(t)g′(t) . · − This is called the center of curvature at P (t). With this point as center, the circle through P is tangent to the curve C. The radius of this circle is 2 2 3 (f (t) + g (t) ) 2 ρ = ′ ′ . f ′(t)g′′(t) f ′′(t)g′(t) −

The curvature at P (t) is κ := 1 . ρ Example 20.1. The normals to a circle all pass through the center of the circle. This center is the center of curvature for every point on the circle. The circle has constant curvature. 219

Example 20.2. For the parabola (x, y)=(at2, 2at), the center of curva- ture of P (t) is the point Q( t). −

P (t)

O F 2a Q(−t)

We may also take (20.1) as deﬁning the envelope of the normal as a parametric curve C∗. This is also called the evolute of C. Thus, the tangents of C∗ are the normals to C, generalizing Proposition 19.1.

x2 y2 Example 20.3. Consider the ellipse a2 + b2 = 1 with parametrization x = a cos t, y = b sin t, where a > b (so that the foci are on the x-axis). The evolute is the parametric curve a2 b2 a2 b2 x = − cos3 t, y = − sin3 t. a · − b ·

Eliminating t, we have

2 2 4 (ax) 3 +(by) 3 = c 3 . This curve is called an astroid. 220 Envelope of normal to a conic

P

O

Q

x2 y2 Example 20.4. For the hyperbola a2 b2 = 1 with parametrization (x, y)=(a sec t, b tan t), the evolute is− the parametric curve a2 + b2 a2 + b2 x = sec3 t, y = tan3 t. a · − b ·

P

O

Q Chapter 21

The cissoid

The cissoid of Diocles was invented to solve the problem of two means proportions: Given two quantities a and b, to ﬁnd x and y such that a, x, y, b are in geometric progression. If b = 2a, this is the problem of duplication of the cube. Given a diameter AB of a circle (O), for every point P on the circle, let P ′ be the reﬂection of P in the diameter perpendicular to AB, and Q the intersection of the line AP and the perpendicular to AB through P ′. The locus of Q is the cissoid.

′ P P

Q

A B X O 222 The cissoid

Consider (O) as the unit circle, A and B the points ( 1, 0) and (1, 0) respectively. If angle ∠P AB = t, then P has coordinates−(cos 2t, sin 2t), and Q is the point 1 cos 2t ( 1, 0) + − (1 + cos 2t, sin 2t)=( cos 2t, 2 sin2 t tan t). − 1 + cos 2t · −

Proposition 21.1. Let the line P ′Q intersect the diameter AB at X. The four quantities XB, XP ′, AX, and XQ are in geometric progression. Proof. The four quantities

XB = 1 + cos 2t = 2 cos2 t,

XP ′ = sin 2t = 2 sin t cos t, AX = 1 cos 2t = 2 sin2 t, − XQ = 2 sin2 t tan t form a geometric progression with common ratio tan t.

Application to the duplication of the cube Let CD be the diameter perpendicular to AB, and M the midpoint of OC. Join B, M to intersect the cissoid at Q. Let X be the intersection of the diameter AB with the perpendicular from Q. Then XB : XQ = OB : OM =2:1, and XP ′ and AX are the two mean proportions between them. In particular, AX = √3 2 OM. ·

′ C P

Q M

A B X O

D 223

MacLaurin’s trisectrix

Given two curves C1, C2, and a pole A, for every line through A inter- secting the curves at P1 and P2, let Q be the point on the line such that AQ = AP1 AP2. The locus of Q is the cissoid with respect to the two given curves.− Consider a circle O(A) and the perpendicular bisector of the radius OA. The cissord with respect to these two curves and with pole A is MacLaurin’s trisectrix.

P1

Q

P2

A O B

To trisect an angle, place it in position BOQ with Q on the trisectrix. Join A, Q to intersect the circle at P1. Then the line OP1 a trisector of the angle.

Proof. Note that triangles OAQ and OP1P2 are congruent. Therefore, OQ = OP2 = AP2 = QP1, and

∠BOP1 = 2∠OP1Q = 2∠QOP1.

This shows that OP1 trisects angle BOQ. 224 The cissoid

Exercise

a (1) Show that the trisectrix has polar equation r = a cos t 4 cos t . (2) Calculate the area enclosed by the loop of the trisectrix.−

Newton’s construction of the cissoid. Let ABCD be a given square, each side of length a. Move an identical square PQRS so that one vertex P slides on the side AD and the side QR always passes through the ﬁxed point B. The locus of the midpoint of P Q is the cissoid. Chapter 22

Conchoids

The conchoids of Nicomedes (2nd century BC) was invented to solve the problem of neuxus, and applied to the trisection of an angle. Given a line ℓ, a point A, and a length b, the chord with axis ℓ and pole A for length b is the locus of a point Q such that when the line AQ intersect ℓ at P , the length of P Q is equal to b.

Q

P

A ′ Q

Let ℓ be the x-axis, A = (0, a) for a > 0. Given a nonzero real number b, we consider the point −Q on the line joining A to P = (t, 0) such that P Q = b. (1) The coordinates of Q are b (x, y)=(t, 0) + (t,a). √a2 + t2 · 226 Conchoids

(2) If we write t = a tan θ, this becomes (x, y)=(a tan θ + b sin θ, b cos θ).

(i) b

Q

O P

a ′ Q θ

A (ii) b = a. | | Q

O

P

′ Q

A (iii) b >a shown above. | | Exercise If b>a, the conchoid with length b has a loop. What is the area enclosed by the loop? −

A 227

This is how the conchoid can be used for the trisection of an angle. Given an acute angle OAB to trisect (assuming OB perpendicular to OA), with A as pole and the line OB as axis, construct the conchoid with length 2 AB. Construct the parallel to OA through B, to intersect the conchoid at· Q. Then the line AQ trisects angle AOB.

Q

C

M

O P B

A

Proof. Join A and Q to intersect the axis at P . Let M be the midpoint of P Q. Since P Q = 2 AB and ∠P BQ is a right angle, we have MQ = MB = AB, and ·

∠MAB = ∠AMB = ∠MBQ + ∠MQB = 2∠MQB = 2∠MAO. This shows that AM trisects angle OAB.

More general conchoids

Let C be a given curve, and A a point, k a ﬁxed length. If P is a varying point on C and on the line AP we mark a point Q such that the directed length P Q = k (positive if P Q is along the same direction of AP ), the locus of Q is called the conchoid CA,k. 228 Conchoids

Conchoids on a circle

The conchoids on a circle are easily described in polar coordinates:

r = a cos θ b. −

(ii) b = a (cardioid): | |

Q

P

A B

′ Q 229

(iii) b >a: | | Q

P

A B

′ Q

Application to trisection of an angle Etienne Pascal (1588-1640) 1 made use of the conchoid with a = 2b to trisect an angle. Q

P

A O B

Exercise (1) Consider the unit circle x2 + y2 = 1. Show that The envelope of the line joining (cos t, sin t) to (cos 2t, sin 2t) is a cardioid.

1Father of Blaise Pascal (1623–1662). 230 Conchoids

(2) Show that for a ﬁxed k, the locus of the point dividing the segment joining (cos t, sin t) to (cos 2t, sin 2t) in the ratio k : 1 k is a conchoid on a circle. −

The witch of Agnesi

Let AB be a diameter with a circle. Given a point P on the circle, join A and P to meet the tangent at B at a point R. The parallel through P and the perpendicular through Q to AB intersect at Q. The locus of Q is the witch of Agnesi. 2 It has parametric equations

x = a cot t, y = a sin2 t.

B a cot t R t

t ? ?

a

a sin t P Q t A

Exercise (1) Locate the points of inﬂection of the witch of Agnesi. (2) Calculate the area under the witch of Agnesi (and above the x- axis).

a

2The infamous name of this curve resulted from a mistake in translation. See [?]. Chapter 23

The quadratrix

The quadratrix of Hippias was invented to solve the problems of angle trisection and quadrature of the circle. This latter means ﬁnding the length a quadrant of a circle in terms of the radius. Consider the quadrant of the circle OAC inside a square OABC. Suppose a radius is rotating uniformly from position OA to OC. At the same time, a segment parallel to AB is moving uniformly from position AB to OC. The locus of the intersection of the rotating radius and the moving segment is the quadratrix.

A B

Q

P X Y

O G C Suppose each side of the square has length a. 2a 2a π (x, y)=( t cot t, t) for 0

Application to angle trisection Suppose an acute angle, given as the arc COQ, is to be trisected. Let the radius OQ intersect the quadratrix at P . Through P construct the segment XY parallel to OC, with X on OA and Y on CB respectively. Trisect the segment CY at Y1 and Y2. The parallels to OC through these points intersect the quadratrix at two points P1 and P2 such that OP1 and OP2 trisect the given angle COP .

A B

Q

P X Y

P2 X2 Y2

P1 X1 Y1

O G C

Quadrature of the circle Let G be the limiting position on the radius OC of the intersection P of the rotating radius and moving segment generating the quadratrix. The AB2 length of the quadrant AQC is equal to OG .