Calculus: Analytic Geometry and Calculus, with Vectors

356 Cones and tonics

a line on the cone. The intersection is a curve of which a part (the solid
part) lies on the front half of the cone and a part (the dotted part) lies on
the back half of the cone. Our present requirement is exceedingly
modest. All we are required to do is grasp the idea that the curve looks
like a parabola. Our solid information about this matter will come in the
next section. Meanwhile, persons with artistic flairs can find useful
entertainment in sketching sections of cones made by planes not parallel
to lines on the cones.

Problems 6.19

1 Sketch a graph showing the parabola whose equation is y = x2 together

with the focus and directrix of the parabola. Draw the focal squares and make

any repairs that may be necessary to make the parabola contain corners of the

focal squares. Prove that the tangents to the parabola at the latter corners

are diagonals of the focal squares, and make any additional repairs that may be

necessary.

2 Problems of this section deal quite exclusively with parabolas placed

upon coordinate systems in such a way that their equations have the standard

form y = kx2, where k is a positive constant. We can, however, pause briefly

to note that the equation

(1) y-yo=k(x-xo)2

is the equation of a parabola having its vertex at the point (xo,yo) Supposing

that a, b, c are constants for which a 0 0, show that the graph of the equation

(2) y=axe+bx+c

is a parabola and find the coordinates of its vertex. Solution: From (2) we obtain

(3) y=a(x2+bxa }+c

z

=a(x2 }

ax+4a2)+c-Tz

a

= a (x +

ya)2+4ac4a bz

and hence J

2 - 4a b 12

(4) Y I- 4a = a(x +Tal

Thus the graph of (2) is a parabola having its vertex at the point (- Za'

b2 - 4ac

4a )- Remark: In case a > 0, the equation (4) has the form (1), where

k > 0 and the graph "opens upward" like the graph of y = kx2. In case a < 0,

(4) has the form

(5) Y-yo=-k(x-xo)2,

where k > 0 and the graph "opens downward" like the graph ofy = -kx2.