MHB Logical error in Spivak's Calculus?

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The discussion centers on a perceived logical error in Spivak's treatment of conic sections, specifically regarding the intersection of a cone and a plane. The user questions the validity of the statement that if the equations for the cone and plane are equal, then a point lies in their intersection, arguing that there could be multiple points satisfying the equation without being on either surface. They highlight that eliminating z leads to a quadratic equation in x and y, which does not guarantee that all points will satisfy the intersection condition. The user concludes that Spivak's exposition is misleading, suggesting that it should clarify the assumption that points must conform to a specific form to validate the proposition. This critique emphasizes the need for precision in mathematical exposition to avoid misinterpretation.
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Yes I am another plucky young fool who decided to self study Spivak. I think I have found an error in his section on conic sections, but Spivak is seldom wrong and I want to be sure I'm thinking straight.

Let $C$ be a cone generated by a line of gradient $m$ which goes through the origin. Then $(x,y,z)$ is on $C$ if $$(1) \qquad z = \pm m \sqrt{ x^{2} + y^{2} }.$$

Let $P$ be a plane which intersects with the cone and whose intersection with the $xy$-plane is a line parallel to the $y-$ plane. Thus, the intersection of $P$ with the $xz$-plane is a line: $L$, say. Supposing $L$ to have gradient $M$ and $z$-intercept $B$, the line $L$ can be described by the equation $$ (2) \qquad z = Mx+B.$$

All is right and well. But then he says 'combining $(1)$ and $(2)$, we see that (x,y,z) is in the intersection of the cone and the plane if and only if $$Mx+B = \pm m \sqrt{ x^{2} + y^{2} }.$$

I understand why, if $(x,y,z)$ is in the intersection, then $(1) = (2),$ but why, is the converse true? Surely we can find an infinite number of points where the equations are equal, but $z$ could be any number and the point not on either plane.

He doesn't first assume that the point is already on $C$ or $P,$ just that it is in $\mathbb{R}^{3},$ and I haven't missed anything in his argument out. Am I just being thick, or do I have a point?
 
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the point is, if you "eliminate $z$" you are left with a quadratic equation of $x,y$ in the $xy$-plane. what happens in the 3 cases:

$M^2-m^2 > 0$
$M^2-m^2 = 0$
$M^2-m^2 < 0$?

not "all points" $(x,y)$ are going to satisfy:

$\{(x,y) \in \Bbb R^2: Mx + B = \pm m\sqrt{x^2 + y^2}\}$

when $M,m,B$ are fixed before-hand.

if we pick such an $(x,y)$, this completely determines $(x,y,Mx+B)$ yes?
 
Thank you for your post. Whilst I don't disagree with anything you have said, I still have a problem accepting the proposition: $$Mx+B = \pm m \sqrt{ x^{2} + y^{2} } \Rightarrow (x,y,z) \in P \cap C. $$ In fact, suppose $(x,y,z) \in P \cap C,$ then by that very proposition it follows that $(x,y,z+1) \in P \cap C$ which, among other things, contradicts the assumption that $C$ is a cone.

I think I have concluded that this is really an error, at least in exposition. He should have made it clear that it is assumed that $(x,y,z)$ has the form $(x,y,Mx+B)$ in which case the proposition definitely holds.
 
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