# [SOLVED]Logical error in Spivak's Calculus?

#### nimon

##### New member
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?

#### Deveno

##### Well-known member
MHB Math Scholar
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?

#### nimon

##### New member
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.