Calculus: Analytic Geometry and Calculus, with Vectors

6.4 Hyperbolas 391

can help usremember, the polar line of P, is the line L, containing thetwo points Q,

and R,. A standard simple proof of this result is as amazing as the result. If the

coordinates of the point Q, on K are (x2,y2), then the equation of the tangent to

K at Qi is

(3) .4x2x + By2y = 1.

The fact that Pi(xl,yl) lies on this line implies that

(4) Ax2x, + By2y, = 1.

This implies that (2) holds when x = x2 and y = y2 and hence that Q, lies on the
line L,. When we have scrutinized this sleight of hand closely enough to under-
stand it, we can see that the same argu-
ment proves that R, is on Ll. Thus we P,
have given geometric interpretationsto
the polar line L, of P, for the case in which
P, lies on K and for the case in which P lies p L,
outside K. Our next geometric inter-
pretation of the polar line L, of P, is equal-
ly applicable to the case in which P, lies
outside the conic (Figure 6.499), the case
in which P, lies on the conic (Figure
6.4991), and the case in which P, lies

Figure 6.499

inside the conic (Figure 6.4992). Let L be a line through P, which intersects the
conic at two points Pz(x2,y2) and Pa(xs,ya). The tangents to the conic at P2 and
Ps have the equations

(5) .4x2x + By2y = 1

(6) .4x3x + By3y = 1.

Except for the special line L which contains the center 0, these tangents intersect

Figure 6.4991 Figure 6.4992

at the point P(x,y) for which the two equations (5) and (6) are both satisfied.

Since P, lies on the line L containing P2 and Pa, there exists a constant A such that

(7) x, = x2 + X(xa - x2) = (1 - X)x2 + Axs

(8) y, = y2 + X(ya - ys) = (1 - X)ys + Xya.