How to derive Kepler's 1st Law?
I must say that I agree with you, this way of derivate the Kepler law is not the most intuitive, it's maybe for this reason they specify: &quot;revisited&quot;.
R1: The reason is stated at the very begin of the chapter, where you derive the relation between $r$ and the angle to the perihelion $\theta$ for the general case of an ellipse (Eq 3 in your book in Cap 2.1 Elliptical orbit, but I can see I have an older version than yours), where:
$$ r = \frac{a(1-e^2)}{1+e\,cos(\theta)}$$
here, the eccentricity of a ellipse is then define as:
$$ e = \sqrt{1- \frac{b^2}{a^2}}$$
in the case of Kepler's law you can define the eccentricity as $e = D/\mu G M$ and  $a$ as:
$$ a = \frac{L^2 G\, M}{\mu^2\,G^2\,M^2 - D^2}$$
and you can work out in equation (29), obtaining the equation in the same form as for an elliptical orbit.
The reason to derive $r$ is because it indicates the distance of the object from the focal point (centre of mass of the system), which in the case of the solar system, is the distance of the planet to the Sun (actually the focal point of the solar system, or better to say of the Sun-Jupiter system, is a few kilometre above the solar surface).
R2: the eccentricity $e$ quantify the shape of a cone cross-section Eccentricity (math) Wikipedia.
For example, in the case of the an ellipse you have $0 &lt; e &lt; 1$, and it indicates how much the ellipse is &quot;squashed&quot;. Another example, if you have $e=0$ than you have a circle.
In the case of $e=1$ you have a parabola, from the equation of $e$ above it means that $a\gg b$. In a more physics sense, it means that the system is not in a gravitationally bounded state, and the less massive object $m_1 \ll m_2$ is on a parabolic trajectory, deviated by the gravitational pull of $m_2$, and will not stay in orbit.