Calculus: Analytic Geometry and Calculus, with Vectors

6.4 Hyperbolas 389

30 Let the angle A10A2 of Figure 6.497 be a
given angle between 0 and a. Everyone start-
ing study of calculus must know that it is very
easy to give aruler-and-compass construction of
the arc A1A2 of a circle having its center at 0
and of the line OX which bisects the angle A10A2.
We may be too busy with other affairs to learn
how the result can be proved, but we should

neverthelessknow that it has beenproved tobe
impossible to give a ruler-and-compass construc-
tion of points P1 and P2 such that the lines OP1

Figure 6.497

and OP2 trisect the given angle. Pappus of Alexandria, who flourished about A.D.
300, trisected the angle by means of hyperbolas. Let H1 be the branch near Al of
the hyperbola having eccentricity 2 for which Al is a focus and OX is a directrix,
and let P, be the point at which H1 intersects the circular arc A1A2. A branch
H2 of a similar hyperbola having a focus at A2 intersects the circular arc at P2.
The definition of eccentricity and the symmetry imply that
I71P11 = 21PiDI = IP1P21 = 21DP2I = IP2A21
Thus the three chords A1P1, P1P2, P2A2 have equal lengths, and it follows that
the two lines OP, and OP2 trisect the given angle A10A2. In modern mathe-
matics, it is important to know why the Pappus construction does not provide a
ruler-and-compass construction of P1 and P2. It is possible to produce hordes
of points on the hyperbolas with rulers and compasses, but it is impossible to
produce all of them. In particular, it is impossible to prescribe rules for ruler-
and-compass construction of the particular points where the hyperbolas intersect
the circular arc. This matter will again be brought to our attention in Problems
10.19.
31 Let P1(x1,y1) be a point on the rectangular hyperbola having the equation
(1) x2-y2 =a2.
In terms of x1 and yl, find the coordinates of the point (x,y) at which the tangent
to the hyperbola at P1 intersects the line through the origin perpendicular to this
tangent. Finally, find an equation which x and y must satisfy by eliminating
x1 and yl from your equations and the equation xi - yi = a2. Ans.: The first
two of the required equations are

(2)

x = a2x1 - aryl

xi +' yi

y

xi +' yi

We can square and add, and then square and subtract, to obtain

(3) x. 2 + y2 = a4 , X a6

X1 + yi (.,2+' y1)2

It is then easy to obtain the final answer

(4) (x2 + y2)2 = a2(x2 - y2).

Remark: The graph of (4) is called the lemniscate of Bernoulli. In Problem 5

of Section 10.1 we shall find that its polar coordinate equation is p2 = a2 cos 20.

Its graph appears in Figure 10.171.