Calculus: Analytic Geometry and Calculus, with Vectors

6.1 Parabolas 357

These matters are important because equations of the form (2) appear very often,

but for basic studies of parabolas we use the standard form y = kx2, where k > 0.

3 Show that the tangent to the graph of the equation y = kx2 at the point

Pl(xi,kxi) has the equation

y - kxi = 2kx1(x - x1) or y = kx,(2x - x1).

Show that this tangent intersects the y axis, the x axis, and the directrix at the

points

.4(0, -kxi),B(-' 0),C(z'- 1,- 1J

(^22) 8k2x1 4k
provided, for the last point, x1 5-1 0. Sketch a figure in which the parabola, the
tangent, and the points B, B, C all appear.
4 Show that the normal to the graph of the equation y = kx2 at the point
Pi(xl,kxi) has, when x1 0 0, the equation
y-kxl=-2kx1(x-xl).^21
Show that this normal intersects the y axis, the x axis, and the directrix at the
points
Sketch a figure showing all of these things.
5 Two points (xi,yi) and (x2,y2) lie on the parabola having the equation
y = kx2. Prove that the coordinates of the intersection R of the y axis and the
line through these points can be put in the form
(0,-kxlx2). Figure 6.191 illustrates results of
this and the next two problems, but the figures
look quite different when x1 and x2 have opposite
signs.
6 Two points (x1,yl) and (x2,y2) lie on the
parabola having the equation y = kx2. Prove
that the coordinates of the intersection of the
tangents to the parabola at these points can be
put in the form
(x, + x2,kxlx2
2
7 Show that the results of the two preceding
Figure 6.191
problems yield the following theorem. Let P1 andP2 be two points on a parabola.
Let Q be the intersection of the tangents to the parabola at P1 and P2. Let R be
the intersection of the line P1P2 and the axis of the parabola. Then the mid-
point of the segment QR lies on the line tangent to the parabola at the vertex.
Solution: The results of the preceding problems show that the mid-point is
Cxl + x2
4 01 , and this point is on the tangent to the parabola at the vertex.