Chapter 26
The parabola
Let F be a given point and a given straight line on a plane. The locus of a variable point P at equal distances from F and is a parabola, with focus F and directrix . If we set up a coordinate system so that F =(a, 0) and is the line x = −a, then the parabola has equation y2 =4ax. Each point on the parabola has coordinates (at2, 2at) for some t.We shall simply call this “the point t of the parabola”.
L
P (at2, 2at)
directrix
−a O F =(a, 0)
We collect some useful information on the parabola. (1) The line joining the points P (t1) and P (t2) has equation
2x − (t1 + t2)y +2at1t2 =0.
This line passes through the focus F if and only if t1t2 = −1. 202 The parabola
(2) Letting t1,t2 → t, we obtain the equation of the tangent at t: x − ty + at2 =0. 1 (− 2 0) It has slope t and intersects the axis of the parabola at at , . (3) The normal at t is the line perpendicular to the tangent at t. It has equation tx + y = at3 +2at. (4) Construction of tangent and normal at t: the circle, center F , passing through t, intersects the axis of the parabola at two points. One of these is on the tangent, and the other on the normal.
t
O F
(5) The area of the triangle with vertices t1, t2, t3 on the parabola is 2 2 1 1 at1 at1 2 2 at2 2at2 1 = −a (t1 − t2)(t2 − t3)(t3 − t1). 2 2 at3 2at3 1
This area is positive or negative according as the triangle t1t2t3 is tra- versed counterclockwise or clockwise. 203
Exercise (1) Use integral calculus to find the area bounded by the parabola and the line joining the points t1 and t2 on the parabola.
t2
O
t1
y2 =4ax
2 (2) The circle through the three points t1, t2, t3 on the parabola y = 4ax intersects it at a fourth point. What is this point? 204 The parabola Chapter 27
Archimedes’ quadrature of the parabola
Archimedes solved the problem of quadrature of the parabola without using integral calculus. He expressed the answer in the following form: Theorem 27.1 (Archimedes). Given two points A and B on a parabola, let C be the point on the parabola the tangent at which is parallel to the 4 line AB. Then the area of the parabola segment AB is 3 of the area of the triangle ABC.
B
C
O
A
y2 =4ax Archimedes called the point C the vertex of the parabolic segment 206 Archimedes’ quadrature of the parabola
AB, and established (Quadrature of the Parabola, Proposition 18) that 2 if A and B are the points t1 and t2 on the parabola y =4ax, then C is t1+t2 the point 2 . Proof. AB 2 t The slope of the line is t1+t2 . Since the tangent at has slope 1 = t1+t2 t , it is parallel to AB if t 2 . This is the point on the parabola through which the parallel to the axis bisects the chord t1t2. In Propositions 19–21, Archimedes showed that if D1 and D2 are the vertices of the parabolic segments AC and CB, then 1 Δ =Δ = · Δ ACD1 CBD2 8 ABC.
B
D2
C
O D1
A
y2 =4ax
2 Δ = − 2( − ) − t1+t2 t1+t2 − = a · ( − Proof. (1) ABC a t1 t2 t2 2 2 t1 4 t2 3 t1) . t1+t2 Δ = 1 ·Δ (2) Replacing t2 by 2 , we obtain ACD1 8 ABC; similarly Δ t1+t2 with CBD2, by replacing t1 by 2 . We paraphrase this with a change of terminology. Instead of calling C the vertex of the parabolic segment, we call it the midpoint of the parabolic arc AB. The above result can be restated as follows. Let C 207 be the midpoint of the parabolic arc AB. By introducing the midpoints of the parabolic arcs AC and CB, we obtain a polygon whose area is 1 4 extra of the triangle ABC. Repeated introduction of the midpoints of 1 parabolic arcs increases the area of the polygon by 4 of the preceding increment. Therefore, the area of the parabolic segment AB is the sum of the geometric progression, beginning with ΔABC, and with common 1 ratio 4 . This is 1 1 2 4 1+ + + ··· Δ = · Δ 4 4 ABC 3 ABC.
Remark. Prior to this, Archimedes had solved the problem in a different way, using geometry and mechanics, detailed in his work The Methods, which was only known in the 19-th century. 208 Archimedes’ quadrature of the parabola Chapter 28
The spiral
A radius of a circle, center O, is rotating as a constant rate. At the same time, a point P is moving on the radius at a constant rate, beginning with O and finishing at the end of the radius when it completes one revolution. The locus of P is a spiral. Archimedes computes the area swept out by the radius in one complete revolution. This is the problem of quadrature of the spiral, with solution in his book On Spirals.
O A
Theorem 28.1 (On Spirals, Proposition 24). The area swept out by the first turn of the spiral is one third of the area of the circle. Proof. Divide the circle into n equal parts and consider the positions of P on the corresponding radii. The area in question is divided into n sectors. A typical one has area between the areas of two circular sectors. 210 The spiral
Specifically for k =1, 2,...,n, the k-th sector has area between two 1 k−1 · circular sector of angle n of the circumference, and radii n R and k · n R. Its area Ak satisfies 2 2 1 k − 1 A 1 k < k < . n n Area of circle n n Therefore, 12 +22 + ···+(n − 1)2 Area of spiral 12 +22 + ···+ n2 < < . n3 Area of circle n3 Archimedes made use of a formula equivalent to 1 12 +22 +32 + ···+ 2 = ( + 1)(2 +1) n 6n n n , which he established as Proposition 10. 1 The above double inequality becomes (n − 1)(2n − 1) Area of spiral (n + 1)(2n +1) < < . 6n2 Area of circle 6n2 1 When n is large, the limits of the two ends are each equal to 3 . Therefore, the area of the spiral is one third of the area of the circle.
1Also, Lemma to Proposition 2 of On Conoids and spheroids. 211
The use of the spiral for angle trisection
P
Q2
P2
P1 Q1
A O To trisect an angle, place it in position AOP with O at the vertex of the spiral, A along the initial radius, and P on the spiral. Divide the segment OP into three equal parts at P1 and P2. With O as center, construct circular arc through P1, P2 to intersect the spiral at Q1 and Q2. The lines OQ1 and OQ2 are the trisectors of angle AOP . 212 The spiral Chapter 30
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?IfP 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: F (t)= 2at(at2 − x)+2a(2at − y) =2a(at3 +2at − tx − y), F (t)= 2a(3at2 +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 3 2a − x 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)=0and 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 30.1. The normal at P (t) to P P ( ) = − − 2 (i) intersects again at P t , where t t t , Q (− ) − t (ii) is tangent to at Q t , and intersects it at Q 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 31
Envelope of normal to a conic
What we have found in §30 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.
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 218 Envelope of normal to a conic 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 neighboring normals, namely,
= ( ) − ( ) · f (t)2+g(t)2 x f t g t f (t)g(t)−f (t)g(t) , 2 2 (31.1) = ( )+ ( ) · f (t) +g (t) 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) ( ) := 1 The curvature at P t is κ ρ . Example 31.1. 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 31.2. 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 (31.1) 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 30.1.
2 x2 + y =1 Example 31.3. Consider the ellipse a2 b2 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,wehave
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
2 x2 − y =1 Example 31.4. For the hyperbola a2 b2 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 221
Generation of the ellipse and hyperbola Consider a fixed point F (c, 0) and a variable point P on the circle x2 + y2 = a2. We find the envelope of the perpendicular to FP at P .IfP has coordinates (a cos θ, a sin θ), then this line has equation F (θ)=0, where F (θ):= (x − a cos θ)(c − a cos θ)+(−a sin θ)(y − a sin θ) =(c − a cos θ)x − a sin θ · y + a(a − c cos θ) = −a(x + c)cosθ − ay · sin θ +(a2 + cx).
To find the envelope of the line, we eliminate θ from F (θ)=0and F (θ)=0.
a(x + c)cosθ + ay · sin θ =(a2 + cx), −a(x + c)sinθ + ay · cos θ =0.
Note that the sum of the squares of the expressions on the left hand side is independent of θ: a2(x + c)2 + a2y2 =(a2 + cx)2.
Collecting terms, we have (a2 − c2)x2 + a2y2 = a2(a2 − c2), or x2 y2 + =1. a2 a2 − c2 Therefore, this is an ellipse if c2
More examples of envelopes Example 31.5. Consider the lines whose intercepts in the first quadrant has a fixed length a. If this line makes an angle t with the (negative) x x + y =1 axis, it has equation a cos t a sin t ,or f(x, y, t):=x sin t + y cos t − a sin t cos t =0. 2 2 Here, ft(x, y, t)=x cos t−y sin t−a(cos t−sin t). Solving f(x, y, t)= 0 and ft(x, y, t)=0simultaneously for x and y,wehave x = a cos3 t, y = a sin3 t. 2 2 2 The envelope is the astroid x 3 + y 3 = a 3 .
Figure 31.1: The astroid
Example 31.6. Consider the family of lines whose intercepts on the x- and y-axes have a fixed total length a. If the x-intercept has length t ≤ a, x + y =1 then the line has equation t a−t ,or f(x, y, t):=(a − t)x + ty − t(a − t)=0.
Here, ft(x, y, t)=−x + y − (a − 2t). Solving f =0and ft =0,we have t2 (a − t)2 x = ,y= . √ √a √ a + = a a This is the parabola x y a. Its focus is the point 2 , 2 , and 2 + =0 t2 (a−t) its directrix is the line x y . The distances from a , a to the 2 2 focus and the directrix are both 2t −√2at+a . 2a 223
F
Figure 31.2: The parabola
Example 31.7. Consider the circle with center (at2, 2at) on the parabola y2 =4ax and passing through the origin. It is given by
f(x, y, t)=x2 + y2 − 2at2x − 4aty.
Here, ft(x, y, t)=−4a(tx + y). Solving f =0and ft =0,wehave −2at2 2at3 (x, y)= , . 1+t2 1+t2 224 Envelope of normal to a conic
Exercise (1) Find the cartesian equation of the envelope. a (2) Consider the circle through the origin and with center at, t on the rectangular hyperbola xy = a2. (i) Find the equation of the circle. (ii) Find the envelope of the circle as a parametrized curve. (iii) Find the Cartesian equation of the envelope. Example 31.8. Consider a point P (t)=(a cos t, b sin t) on the ellipse 2 x2 + y =1 ( ) a2 b2 . The equation of the circle with center P t and passing through the origin is x2 + y2 − 2a cos t · x − 2b sin t · y =0
To find the envelope, we eliminate t from 1 cos + sin = ( 2 + 2) ax t by t 2 x y , −ax sin t + by cos t =0 by noting that the sum of the squares of the expressions on the left hand side is independent of t: 1 2 2 + 2 2 = ( 2 + 2)2 a x b y 4 x y .
Exercise Find the polar equation of the curve.