Line $L_t\subset\mathbb{P^2}$ contains $(t, 0)$ and $(0, t + m)$. Show collection of $L_t$ form points of a conic $K^*\subset\mathbb{P}^{2*}$ .

Question

Let $m\ne 0$ be a nonzero real number. For each $t\in\mathbb{R}$, let $L_t\subset\mathbb{P^2}$ be the line containing the points $(t, 0)$ and $(0, t + m)$ in the finite plane. $\mathbb{P^{2*}}$ is $\mathbb{P^{2}}$ dual. $K^*$ is $K$ dual.

Show that the collection of lines $L_t$ form (all but finitely many) points of a conic $K^*\subset\mathbb{P}^{2*}$ and find its equation. What additional lines do you have to add to the family $\{L_t\}_{t\in\mathbb{R}}$ to describe all the points of $K^*$?

My attempt:

$L_t:(t+m)x+ty-t(t+m)=0$ in the finite plane.

Homogeneize: $(t+m)x+ty-t(t+m)z=0$

So the collection of lines $L_t$ is the collection of points $[t+m:t:-t(t+m)]$ in $P^{2*}$. However I cannot find such a conic $K^*\subset \mathbb{P^{2*}}$. I only got $xy=z$, but this is not a homogeneous polynomial. So, I totally have no idea how to do this problem.

Answer 1

You've made good a good start on the problem, and I think you just made an error in your homogenization. Let's work in an affine patch until the very end. You found the line $$ L_t: (t+m)x + ty - t(t+m) = 0 \quad \text{or equivalently} \quad L_t: -\frac{1}{t}x - \frac{1}{t+m}y + 1 = 0 $$ for $t \neq 0, -m$. This corresponds to the point $[X^*, Y^*, Z^*] = \left[-\frac{1}{t}:-\frac{1}{t+m} : 1\right]$ in $(\mathbb{P}^2)^*$. Working in the affine patch where $Z^* \neq 0$ and letting $x^* = X^*/Z^*$ and $y^* = Y^*/Z^*$, then we have $x^* = -\frac{1}{t}$ and $y^* = -\frac{1}{t+m}$. Solving for $t$, we obtain $$ -\frac{1}{x^*} = t = -\frac{1}{y^*} - m $$ and multiplying through by $x^* y^*$ yields $$ -y^* = -x^* - mx^* y^* \implies x^* - y^* + m x^* y^* = 0 \, . $$ This is the conic $K^*$, minus the finitely many points where $Z^* = 0$. To find the full conic we rehomogenize : $$ 0 = x^* - y^* + m x^* y^* = \frac{X^*}{Z^*} - \frac{Y^*}{Z^*}+m\frac{X^*}{Z^*}\frac{Y^*}{Z^*} \, . $$ Multiplying through by ${Z^*}^2$ to clear denominators includes the points at infinity where $Z^* = 0$, which yields the conic $K: X^* Z^* - Y^* Z^* + m X^* Y^* = 0$.

The place you went astray was interpreting the expression $[t+m:t:-t(t+m)]$. These polynomials are not homogeneous of the same degree. We should really working $\mathbb{P}^1$ with coordinates $S,T$, where we consider $\mathbb{R}$ as the affine open subset $U \subseteq \mathbb{P}^1$ where $T \neq 0$ with affine coordinate $t = S/T$. Then the line $L_t = L_{[S:T]}$ corresponds to the point $$ [t+m:t:-t(t+m)] = \left[\frac{S}{T} + m : \frac{S}{T} : -\frac{S}{T}\left(\frac{S}{T}+m\right) \right] = [ST + mT^2 : ST : -S^2 - STm] \, . $$

Answer 2

HINT: Perhaps it would help to put the original points in $\Bbb P^2$ to start with. Thinking of $t\in\Bbb R\subset\Bbb P^1$ with homogeneous coordinates $[t:u]$, we have the points $[t:0:u]$ and $[0:t+mu:u]$. Now our corresponding line is given by $[u(t+mu):tu:-t(t+mu)]\in\Bbb P^{2*}$. Note that these are homogeneous of degree $2$ in the coordinates on $\Bbb P^1$, and so the image is indeed a conic. Here's its equation (remember that $m\in\Bbb R$ is a fixed constant): $$(x-y)(-z-my)=my^2 \iff (x-y)z+mxy=0.$$ (Here, of course, as you were thinking, $x=u(t+mu)$, $y=tu$, $z=-t(t+mu)$.)

Now I leave it to you to find the points on that conic that are actually missing from $\{L_t\}$.