I have a problem which, once again, is probably quite simple, but I can't get my head around how I'm supposed to show the required equivalence. The following problem is from a book called 'Linear Algebra', and the problem is within a sub-chapter where rotating the coordinate axes has just been introduced so using analysis (or anything not relating to linear algebra) is not preferred. Here's the problem:
Suppose that $x_0$ is a point on the genuine conic section $\alpha$: $x^\top Ax + b^\top x + c = 0$. (Here $A = A^\top \in \Bbb{R}^{2 \times 2}$, $b \in \Bbb{R}^2$ and $c \in \Bbb{R}$.) Show that the line $x = x_0 + tu$ (where $t \in \Bbb{R}$) is tangent to $\alpha$ if and only if $u$ is perpendicular to the vector $2Ax_0 + b$.
I first assumed that $u$ is perpendicular to the vector $2Ax_0 + b$. This means that their inner product is zero. Then, by substituting the line's equation to the conic section's equation gives
\begin{align*} &(x_0 + tu)^\top A(x_0 + tu) + b^\top(x_0 + tu) + c = 0 \\ \iff \, &x_0^\top Ax_0 + b^\top x_0 + c + tu^\top(2Ax_0 + b) + t^2u^\top Au = t^2 u^\top Au = 0. \end{align*}
These were used above:
- $x^\top_0 Au = u^\top Ax_0$,
- $u^\top(2Ax_0+b)$ was assumed to be equal to $0$,
- $x_0^\top Ax_0 + b^\top x_0 + c = 0$ because it was assumed that $x_0 \in \alpha$.
But there, at the end, I'm not sure if I can just assume that $u^\top Au \neq 0$. Because If I can, then $t^2$ must be equal to $0$, which implies that the line $x = x_0+tu$ is a tangent to the conic section because then $x_0$ (being any point on $\alpha$) is the only point of the line that touches $\alpha$.
To show the other direction I have no idea how I could achieve this. Nothing seems to work with the tools I have available. Can you give me a hint or two about how to proceed?
Edit 0: Grammar and typos.
Edit 1: Here, $x_0$ is a fixed vector of $\Bbb{R}^2$.
Edit 2: Genuine conic section means that the curve is either a real ellipse, a hyperbola, or a parabola (not a degenerate version of any of them).
We need to be careful here - it is certainly possible for a line to intersect a conic section at only one point and yet not be tangent (consider the $y$-axis for $y=x^2$.)
On that note, we need to clarify what it means for a line to be tangent to a conic section. If you wish to avoid analysis, I would propose we say that $x^0+tu$ is tangent to $\alpha$ if $x^0$ is on $\alpha$ and the line does not cross $\alpha$, which we might formalize by demanding one (and only one) of the following inequalities is simultaneously satisfied for all $t \neq 0$ and $z=x^0+tu$:
$$z^\top A z + b^\top z + c >0\text{,}$$ $$z^\top A z + b^\top z + c <0\text{.}$$
Now, using calculations similar to the ones in your question, we obtain
\begin{align*} &(x_0 + tu)^\top A(x_0 + tu) + b^\top(x_0 + tu) + c \\ = \, &x_0^\top Ax_0 + b^\top x_0 + c + tu^\top(2Ax_0 + b) + t^2u^\top Au \\ = \, & tu^\top(2Ax_0 + b) + t^2 u^\top Au \text{.} \end{align*}
The last expression is a quadratic in the variable $t$ (unless $u^\top Au=0$, which we will deal with separately), and so the vertex is at $t=-u^\top(2Ax_0 + b)/(2 u^\top Au)$. The line is tangent to $\alpha$ if and only if the quadratic equals $0$ only at $t=0$ and does not change sign, or equivalently, the vertex is at $t=0$, which in turn occurs if and only if $u^\top(2Ax_0 + b)=0$.
Finally, we deal with the case where $u^\top Au=0$, in which case the aforementioned "quadratic" is not a quadratic at all, but simply a non-constant* linear function $ tu^\top(2Ax_0 + b)$, which passes through $0$ at $t=0$ alone, and thus $u^\top (2Ax_0+b)$ is nonzero. Since the function is linear and non-constant, it must take on positive and negative values, and thus the line is not tangent per our definition.
(*Note that we know it is non-constant since otherwise, it would be constantly equal to $0$, and then the entire line would pass through the conic section, contradicting nondegeneracy.)