history of mathematics - National STEM Centre

N-
i-
L
Descartes
Thus I have
In the same way, if I had
x = -ax + b ,
lia'+b*.
PM would be x 2 and I should have
and so for other cases.
Activity 7.4 Reflecting on Descartes, 9
Figure 7.6
^ b
R
-o
Q
M
1 In the case of the equation x 4 = -ax 2 + b 2 , PM will be equal to x 2 (Figure 7.5).
Show how you can construct x from this result.
2 What can you say about the law of homogeneity in this case?
To continue:
16 Finally, if I have z 2 = az - b 2 , 1 make NL equal to ja and LM equal to
b as before; then, instead of joining the points M and N, I draw MQR
parallel to LN, and with N as a centre describe a circle through L cutting
MQR in the points Q and R; then z, the line sought, is either MQ or MR,
for in this case it can be expressed in two ways, namely:
and
z = ja-^a -b .
f Activity 7.5 Reflecting on Descartes, 10
90
17 And if the circle described aboutN and passing through L neither
cuts nor touches the line MQR, the equation has no root, so that we may
say that the construction of the problem is impossible.
Figure 7.6 illustrates passages 16 and 17 above. The circle with centre N and radius
ja is drawn. NL is parallel to MR, and O is the mid-point of QR.
1 Show that NO is parallel to LM.
2 Express OQ in terms of a and b.
3 Express MQ and MR in terms of a and b.
4 Why does Descartes consider two solutions in this text?
5 In which case can a solution not be found? In which paragraph does Descartes
describe this?