- #1
nimon
- 3
- 0
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?
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?