Calculus: Analytic Geometry and Calculus, with Vectors

398 Cones and conics

the graph must be a conic. While it is possible to foresee some of the resultsof
putting (2) into standard form by "rotation of axes," we examine the details of
the process. Show that substituting

(3) x=x'cos0-y'sin0, y= x' sin 0 + y' cos 0

into (2) gives an equation in x' and y' for which the coefficient of x'y' will be zero

when tan 20 = -1 and hence

(4) 20= 4, 0=3g, cos20=1+c2

s20-V-12-

1 -cos2B -1

sine B = 2 -

2 /

Continue the work to show that the equation in x' and y' can be put in the form

x'2 y'2

(5) 2-\/2- +2 2N/2- -2=1.

Note that the result agrees very well with Figure 6.591. The distance from the

center of the hyperbola to its vertices is -2V2-. It is possible to use (5)

to obtain more information in primed coordinates and to express this information

in terms of the original coordinates, but all of these operations consume consid-

erable time.

8 Obtain preliminary information about the graph of the equation

x(x+y)=1

by solving for y (or for x) and making a rough sketch. Then obtain more precise

information by rotation of axes.

9 Apply the procedure of Problem 8 to the equation

x2 + xY + y2 = 1.

10 Make and solve more problems similar to Problems 8 and 9, but keep

the equations simple. We still do not have time to do chores that computers

should do.

11 When a and b are positive, the equation

a2x2 + b2y2 - a'--b2 = 0

has elliptic type. For what values of k does the equation

a2x2 + kxy + b2y2 - a2b2 = 0

have elliptic type? Parabolic type? Hyperbolic type? Hint: The discriminant

is kt - 4a2b2, and we are put on the right track when we notice that this is nega-

tive and our equation is elliptic when k = 0.

12 When a and b are positive, the equation

a2x2 - b2y2 - a2b2 = 0

has hyperbolic type. For what values of k does the equation

a2x2 + kxy - b5y2 - a2b2 = 0