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Apollonius' Extremum Problems on Conics

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Apollonius' Extremum Problems on Conics 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.
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