Calculus: Analytic Geometry and Calculus, with Vectors

(lu) #1

26 Analytic geometry in two dimensions


When the parentheses are removed and the constant terms are collected,
this equation takes the less informative form

(1.452) x2 + y2 + 4x - 6y - 12 = 0.

This has the form

(1.453) x2+y2+Dx+Ey+F=O,


where D, E, and F are constants. It turns out that for some sets of
values of D, E, and F, (1.453) is the equation of a circle. To try to write
(1.453) in the standard form (1.44), we begin by writing it in the form

(1.454) (x2 + Dx + ) + (y2 + Ey + ) = -F.


The next step is to add a constant to the term x2 + Dx so that the sum
will be the square of a quantity of the form (x + Q). What shall we add?
A good look at the formula

(x+Q)2=x2+2Qx+Q2


provides the answer: divide the coefficient of x by 2 and square the result.
Thus we add D2/4 and E2/4 to both sides of (1.454) to obtain

or

(1.46) x+DI2+(y+2)2=D2+

E2

We can now see how the graph depends upon the constants D, E, and F.
In case D2 + E2 - 4F > 0, then (1.46) is the equation of the circle with
center at (-JD, ---E) and radius D2 + E2 - 4F. In case
D2 + E2 - 4F = 0, the equation becomes

(1.461) (x + -D)2 + (y + JE) 2 = 0.

This equation is satisfied when and only when x = -jD and y = -4E
so the graph is the single point (-JD, --E). In case D2 + E2 - 4F < 0,
there are no pairs of values of x and y for which the equation is satisfied.
One is tempted to say that the poor equation has no graph, but the graph
is actually the empty set, that is, the set havingno points in it. Thus,
determination of the graph of the equation

x2+y2+6x-7y+8 =0


is made by completing squares. The process is important and must be
remembered.
Before starting the next paragraph, we look atsome algebra and ways
in which it is printed. The quotient a/bc is called a shilling quotient and
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