Find the intersection of a cone and its tangent plane.
My partial solution is below. Please help me complete it.
Note: Solutions exist. This question asks to help me complete my solution, which uses vector methods.
For any point $v$ in the plane, let $c(v) : \mathbb R^2 \to \mathbb R$ be the non-negative $z$-coordinate on the cone for $v$. Then $$c(v) = \|v\|$$ and $$Dc(v) = \frac {v^\top} {\|v\|}.$$
Fix $a$ in the plane, and let $T_a$ be the plane tangent to the cone at $(a, \|a\|)$. For any point $v$ in the plane, let $t_a(v): \mathbb R^2 \to \mathbb R$ be the $z$-coordinate over $v$ of $T_a$. Then $$\begin{align*} t_a(v) &= c(v) + Dc(a)(v-a) \\ &= \|v\| + \frac{v^\top (v-a)}{\|v\|} \\ &= \|v\| + \frac {v} {\|v\|} \cdot (v-a). \end{align*}$$
For $z \geq 0$, the cone and $T_a$ intersect at $(v, z)$ iff $z = t_a(v) = c(v)$, which is when $v \cdot (v-a) = 0$ and $z = \|v\|$. If $a = 0$, this is the point at the origin.
Otherwise (for $a \neq 0$), I believe this will be a ray through the origin, and by symmetry the intersection below (for $z < 0$) will extend the ray to a line through the origin - but am having trouble completing the proof. (See also this question).
Update
What I wrote above is incorrect. Thank you to Ted Shifrin for helping me find this mistake! Rather, it should be
$$\begin{align*} t_a(v) &= c(a) + Dc(a)(v-a) \quad \text{(Here was my error!)}\\ &= \|a\| + \frac{v^\top (v-a)}{\|v\|} \\ &= \|a\| + \frac {v} {\|v\|} \cdot (v-a) \\ \end{align*}$$
So the projection of the intersection onto the $xy$ plane is the locus of points $v$ such that $\|v\| = \|a\| + \frac {v} {\|v\|} \cdot (v-a)$, and the intersection is the set $\{(v, \|v\|)\}$. I'm still struggling to prove this is a line through the origin.
