history of mathematics - National STEM Centre

3 Yes. When the pin A, for example, moves along the
*-axis, the straight line motion of pin B is directly linked
to it. The curve traced out by P is directly linked to these
two preceding motions.
Activity 7.12, page 96
1 Take, for instance, EG as the positive jc-axis and EC
as the positive ^-axis. FD intersects BG in R, as shown in
the figure.
Suppose that EG = x , EC - y and EB = p .
BG is perpendicular to FD, and triangle BEG is similar to
triangle GRD. From this you can deduce that
Mp 2 + x 2 } = py. This is the equation of a parabola
because you can write it in the form x2 = 2p(y -
3 This instrument satisfies the conditions. The motion
of FK is directly linked to the straight line motion of G,
and the motion of D is in turn directly linked to the
motion of the ruler FK, and of GP.
Activity 7. 13, page 98
I Triangles BYC and C YD are similar so — = —— , and
, , , y BC
from triangle BYC, BC 2 = x-a.
Activity 7.14, page 99
1 He forgot the straight line.
Activity 7.15, page 99
x 2 + y 2 + qx + 4y = 0
2 The x-values of g, G and F are the points of
intersection of x 2 +y 2 + qx + 4y = 0 and y = -x 2 .
Therefore, substituting y = -x 2 , you find
x 2 + x 4 + qx - 4x 2 = 0 , that is, jc 4 - 3x 2 + qx = 0 or
14 Answers
x(x 3 - 3x + q\ - 0. As x = 0 is not a suitable solution,
you are left with x 3 — 3x + q = 0 .
Activity 8.2, page 102
1 The degree of the equation of the curve defines the
class of the curve.
Activity 8.3, page 103
1 'When the relation between all points of a curve and
all points of a straight line is known, in the way I have
already explained, it is easy to find the relation between
the points of the curve and all other given points and
lines.'
Activity 8.4, page 103
1 To find the angle formed by two intersecting curves.
Activity 8.5, page 104
Activity 8.6, page 105
1 He calls these negative solutions 'false roots'.
2 The defect of a quantity 5, is the number -5.
Activity 8.7, page 105
1 One positive solution; two positive solutions
2 See the remarks immediately following the activity.
Activity 8.8, page 106
1 The equation has two positive solutions. The sign rule
works.
2 Two changes of sign, so two positive solutions.
3 The .^-coordinate of F.
Activity 8.9, page 108
I The equation can be written in the form
.
2 I
2 a The circle has centre (/7,0) and radius
3 - p) + -f . Its equation is therefore
(x — p) + y =(3- p) + -| which simplifies to
r = -*
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