Empty real conic containing two pairs of conjugate points in the projective plane? Given two conjugate pairs of points in general position in $\mathbb{CP}^2$, there is a pencil of real conics containing these four points. Is there a real empty conic in this pencil? 
 A: Yes, there is a real empty conic containing the points.  Denote the first conjugate set of points by $\{p,\overline{p}\}$, and denote the second conjugate set of points by $\{q,\overline{q}\}$.  If the pairs are "general", then no three of these four points are collinear.  In particular, the lines $L=\text{span}(p,q)$ and $\overline{L}=\text{span}(\overline{p},\overline{q})$ are distinct and intersect in a unique real point $r$.  The union $L\cup \overline{L}$ is invariant under conjugation, hence it has a (homogeneous, degree 2) defining polynomial $F$ that with real coefficients.  Choose a real chart $B$ about $r$ (or rather about a preimage point in $\mathbb{R}^3\setminus\{0\}$, or alternatively, dehomogenize $F$ on $B$).  Since the only zero of $F$ on $B$ is at $r$, either $F$ is nonnegative with unique zero at $r$, or $F$ is nonpositive with unique zero at $r$.  Up to replacing $F$ by $-F$, assume that $F$ is nonnegative.
Now consider the lines $P=\text{span}(p,\overline{p})$ and $Q=\text{span}(q,\overline{q})$.  If $r$ were contained in $P$, then $P$ would equal $L$ so that $q$ is contained in $P$, i.e., $p$, $\overline{p}$ and $q$ are collinear, contrary to the hypothesis that the pairs are "general".  Thus $r$ is not contained in $P$.  By the same argument, also $r$ is not contained in $Q$.  Thus $r$ is not contained in $P\cup Q$.  Since $P\cup Q$ is invariant under conjugation, there is a defining equation $G$ with real coefficients.  Since $r$ is not contained in $G$, either $G$ is positive or negative on $r$.  Up to replacing by $-G$, assume that $G$ is positive.  Up to shrinking $B$, we may assume that $G$ is positive on all of $B$.  
Since $F$ has only one zero in $\mathbb{R}P^2$ at $r$, and since $G$ is positive on a neighborhood $B$ of $r$, for sufficiently small $\epsilon>0$, $F+\epsilon G$ has no zeroes in $\mathbb{R}P^2$ (I am using that $\mathbb{R}P^2$ is compact).
