I have a question regarding the conics section of Spivak calculus and the way he derives the final equation. I understand everything that he is doing up until:
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 $\beta$
Now, I understand everything before that, and why he is changing coordinates from the 'standard' ones, xyz, to coordinates within the plane, so he can express the intersection of the cone and the plane in a equation. But the thing is how does he justify that there is some $\alpha$ and some $\beta$ that satisfy this condition for all x ? He just claims that such values exist, with no proof.
For anyone not familiarized with this chapter, i leave it down here so you can read it.
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.
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).
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.
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.
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
$$C^2y^2+(C^2-M^2)x^2+Gx+H=0.$$
Best Answer
You're correct that his presentation is a bit confusing. He's using $x$ both for the original coordinate (in the $xyz$ coordinate system) and to represent the new coordinate along $L$ in the plane $P$. So to clarify, let's use $x'$ as the indicated (signed) distance from the point $Q$ to the point (which I'll call $R$ for now) on $L$. Then Spivak's claim is that the original $xz$-coordinates of $R$ are linear expressions in $x'$. Let $S$ be the point on the $x$-axis directly below $R$.
If you'll allow me to introduce a little trigonometry for a moment, let $\theta$ be the angle between the two lines, and let $Q=(\beta,0)$. Then $\alpha=\cos\theta = QS/QR$, so $QS = \alpha(QR) = \alpha x'$. Since $S=(x,0)$, we have $x-\beta = \alpha x'$, so $x=\alpha x' + \beta$. Of course, we do not need the definition of the cosine function to observe that as $R$ moves along the line $L$, the ratio of the signed distances $QS/QR$ stays constant; this is just similar triangles.
Now proceed with his argument, substituting $x=\alpha x'+\beta$. The $x'y$-coordinates in the plane $P$ are, of course, not the usual $xy$-coordinates coming from $3$-space.
Last comment: Since the equation of $L$ is $z=Mx+B$, then we have $\beta = -B/M$ and $M=\tan\theta$, so $\alpha=\cos\theta = (\pm)\frac1{\sqrt{M^2+1}}$.