Calculus: Analytic Geometry and Calculus, with Vectors

6.2 Geometry of cones and conics 367

This equation is ponderous and nobody should ever dream of remembering it,

but it is useful.

2 With the aid of the result of Problem 1, show that the coefficient ofxy

in the equation of a conic K is zero if and only if each directrix of K is parallel

to one of the coordinate axes.

3 With the aid of the result of Problem 1, show that if the coefficients of

x2 and y2 in the equation of a conic K are both 0, then e = - and 42= B2.

Show that when e = - and B = 11, the equation reduces to

(2)

C)Y=_2 i y1- C2

rl 2 2112

Show that when e = - and B = -A, the equation reduces to

(3)

C) C) C2 X2 + Y2

xy

x1 X11

X-

YI-rl (^5) 2,42 2
The graphs of these equations are hyperbolas because e = \ > 1.
4 With the aid of Problem 3, show that if e = N/"2-, B= .4, x1 = -C/A,
and yl = -C/A, then the equation of the conic K is
(4)
XY = C2
Show also that if e = -\/2, B = -11, x1 = -C/!1, andy1 = C/fl, then the equa-
tion of the conic K is
(5) xY = - 212.C2
Our equations are now much simpler.
5 Let k > 0. Observe that formula (4) of Problem 4 reduces to xy = k
when 4 = 1 and C = -. This shows that xy = k is the equation of the
conic having a focus at the point ( 25), having a directrix with the
equation
(6) x -I- y - = 0,
and having eccentricity e = N/2_. Remark: Relatively few persons have enough
courage to undertake to sketch or otherwise describe a cone in Es which intersects
the xy plane in the graph of the equation xy = k and then use methods of syn-
thetic geometry (geometry which, unlike analytic geometry, never uses algebra
and other brands of mathematical analysis) to obtain information about the foci
and directrices of the conic. However, the results of this problem enable us to
put this information in very simple terms. Supposing that k > 0, we start with
a good clean x, y coordinate system and show how to locate the foci and directrices
of the hyperbola having the equationxy = k. The point V(NII, 1/k) clearly