Here is a section from the conic sections appendix (Appendix 2) of Spivak's Calculus where I think there is an error:
One of the simplest subsets of this three-dimensional space is the (infinite) cone illustrated in Figure 2; this cone may be produced by rotating a "generaling line," of slope $C$ say, around the third axis.
https://i.stack.imgur.com/tIsW6.png
For any given first two coordinates $x$ and $y$, the point $(x,y,0)$ in the horizontal plane has distance $\sqrt{x^2+y^2}$ from the origin, and thus $$\tag1 (x,y,z)\text{ is in the cone if and only if }z=\pm C\sqrt{x^2+y^2}.$$ We can descend from these three-dimensional vistas to the more familiar two-dimnensional one by asking what happens when we intersect this cone with some plane $P$ (Figure 3).
https://i.stack.imgur.com/epVDr.png
If the plane is parallel to the horizontal plane, there's certainly no mystery--the intersection is just a circle. Otherwise, the plane $P$ intersects the horizontal plane in a straight line. We can make things a lot simpler for ourselves if we rotate everything so that this intersection line points straight out from the plane of the paper, while the first axis is in the usual position that we are familiar with. The plane $P$ is thus viewed "straight on," so that all we see (Figure 4) is its intersection $L$ with the plane of the first and third axes; from this view-point the cone itself simply appears as two straight lines.
https://i.stack.imgur.com/nQSRP.png
In the plane of the first and third axes, the line $L$ can be described as the collection of all points of the form $$(x,Mx+B),$$ where $M$ is the slope of $L$. For an arbitrary point $(x,y,z)$ it follows that $$\tag2 (x,y,z)\text{ is in the plane }P\text{ if and only if }z=Mx+B.$$ Combining $(1)$ and $(2)$, we see that $(x,y,z)$ is in the intersection of the cone and the plane if and only if $$\tag{$*$} Mx+B=\pm C\sqrt{x^2+y^2}.$$ Now we have to choose coordinate axes in the plane $P$. We can choose $L$ as the first axis, measuring distances from the intersection $Q$ with the horizontal plane (Figure 5); for the second axis we just choose the line through $Q$ parallel to our original second axis. If the first coordinate of a point in $P$ with respect to these axis is $x$, then the first coordinate of this point with respect to the original axes can be written in the form $$\alpha x+\beta$$ for some $\alpha$ and $\beta$. On the other hand, if the second coordinate of the point with respect to these axes is $y$, then $y$ is also the second coordinate with respect to the original axes.
https://i.stack.imgur.com/iWT2a.png
Consequently, $(*)$ says that the point lies on the intersection of the plane and the cone if and only if $$M(\alpha x+\beta)+B=\pm C\sqrt{(\alpha x+\beta)^2+y^2}.$$ Although this looks fairly complicated, after squaring we can write this as $$\alpha^2C^2y^2+\alpha^2(M^2-A^2)x^2+Ex+F=0$$ for some $E$ and $F$ that we won't bother writing out. Dividing by $\alpha^2$ simplifies this to $$C^2y^2+(C^2-M^2)x^2+Gx+H=0.$$
Here is where I think is an error:
He squares $$M(\alpha x+\beta)+B=\pm C\sqrt{(\alpha x+\beta)^2+y^2}$$ to get $$\alpha^2C^2y^2+\alpha^2(M^2-A^2)x^2+Ex+F=0$$ Shouldn't it be $$\alpha^2C^2y^2+\alpha^2(C^2-M^2)x^2+Ex+F=0$$
And what does the value of $G$ and $H$? They are irrelevant for the type of the conic section. What do they affect?