Commentary on Proposition X.49

Back to Proposition X.49.

[Folio 0109 | p. 105]

Taking a rational straight line $kp$ commensurable in length with $\rho$, and selecting numbers of the same form as before, viz. $p(m^2 - n^2)$, $pn^2$, we put

$$p(m^2 - n^2) : pm^2 = k^2 p^2 : x^2 \dots\dots\dots\dots\dots\dots\dots(1),$$

so that

$$x = kp \frac{m}{\sqrt{m^2 - n^2}}$$

$$= kp \frac{1}{\sqrt{1 - \lambda^2}}, \text{ say} \dots\dots\dots\dots\dots\dots\dots\dots(2).$$

Just as before, $x$ is rational and $\sim kp$, whence $kp + x$ is a binomial straight line.

By (1),

$$x^2 > k^2 p^2.$$

[Folio 0110 | p. 106]

Let

$$x^2 - k^2 p^2 = y^2,$$

whence, from (1),

$$pm^2 : pn^2 = x^2 : y^2,$$

and $y$ is therefore rational and $\sim x$.

The greater term of the binomial straight line is $x$ and the lesser $kp$, and

$$\frac{kp}{\sqrt{1 - \lambda^2}} + kp$$

satisfies the definition of the second binomial straight line.

The corresponding second apotome [x. 86] is

$$\frac{kp}{\sqrt{1 - \lambda^2}} - kp.$$

The equation of which the two expressions are the roots is

$$x^2 - \frac{2kp}{\sqrt{1 - \lambda^2}} \cdot x + \frac{\lambda^2}{1 - \lambda^2} k^2 p^2 = 0,$$

or

$$x^2 - 2ax + \lambda^2 a^2 = 0,$$

where

$$a = \frac{kp}{\sqrt{1 - \lambda^2}}.$$