Calculus: Analytic Geometry and Calculus, with Vectors

6.5 Translation and rotation of axes 397

of the focus F and the primed equation of the directrix D of the parabola. Finally,

find unprimed equations and coordinates of the parabola and of A, B, F, and D.

2 Write the equation

(1)

in the form

(2)

Then substitute

(3)

(x +

2)2

+(y 2

3)2

z =1

4x2+By2+Cx+Dy+E=0.

x=x'+h, y=y'+k

into the result and so determine h and k that the coefficients of x' and y' will be
zero. Show that the new equation can then be put in the form

x'2 y'2

(4) 12 +22 =1.

Finally, sketch properly related unprimed and primed axes and sketch the graph

of (4).

3 Show how a primed coordinate system can be introduced to simplify the

process of obtaining basic information about the hyperbola having the equation

(x-3)2_(y-1)2=1.

32 12

Sketch a figure showing both sets of axes and the hyperbola.

4 Translate axes to remove the first-degree terms from the equation

xy - 2x + 3y - 4 = 0.

Remark: This gibberish means something. Let x = x' + h, y = y' + k and

determine h and k in such a way that the coefficients of x' and y' in the new equa-

tion will be 0. Ans.: h = -3, k = 2, x'y' = -2.

5 Translate axes to simplify the equation

x2+xy+y=3.

Ans.: Putting x = x' + h and y = y' + k, we find that the first-degree terms

disappear (have zero coefficients) when h = -1, k = 2, and that the simplified

equation is x'2 + x'y' = 2.

6 Construct and solve more problems Figure 6.591

similar to the preceding two, but keep the

equations simple. We do not have time to

do chores that computers should do.

7 A rough graph of the equation

(1) y=x+z

is easily drawn. A miniature version appears

in Figure 6.591. Since (1) can be put in the

form

(2) x2-xy+1 =0,

x