Calculus: Analytic Geometry and Calculus, with Vectors

28 Analytic geometry in two dimensions

This holds if and only if

x2+iy 4k12=(y+4k)2.

Simplifying this gives the very simple and attractive equation

(1.472) y = kx2.

This is the equation of the parabola shown in Figure 1.47.

The point V on a parabola which lies midway between the focus and

directrix of the parabola is called the vertex of the parabola. For example,

when k ; 0, the point (xo,yo) is the vertex of the parabola for which the

point F(xo, yo + 1/4k) is the focus and the line L having the equation

y = yo - 1/4k is the directrix. As Problem 26 invites us to discover, the

equation of this parabola is

(1.473) y - yo = k(x - xo)2.

When k 76 0, the equation

(1.474) y = kx2 + ax + b

can be put in the form (1.473) by completing a square and transposing.

Thus, when k 94- 0, the graph of (1.474) is a parabola, and we must always

remember the fact. The distance from the focus to the vertex isIWhen^1

the positive y axis lies above the origin as it usually does, the focus

is above the vertex when k > 0 and is below the vertex when k < 0.

It is possible to proceed in various ways to calculate the distance

d from a given point P(xo,yo) to the line L having the given equation

Ax + By + C = 0. Problem 34 at the end of this section requires that

the answer be worked out in a specified straightforward way. It is some-

times convenient to omit the calculations and use the result, which is set

forth in the following theorem.

Theorem 1.48 The distance d from the pointP(xo,yo) to the lineL having

the equation Ax + By + C = 0 is given by the formula

(1.481) d =

jJxo+Byo+C1

vl-,41 _+B2

In some applications of this theorem, we use the version obtained by

deleting the subscripts.

Problems 1.49

(^1) Draw the triangle having vertices at the points d(2,2), B(-5,-2), and
C(-2,-4). Calculate the lengths a, b, and c of the three sides BC, CA, and .4B
and show that c2 = a2 + V. This implies that the triangle is a right triangle