Hyperoval

Remarks

Every line meets a hyperoval $H$ in $0$ or $2$ points. Suppose a line met $H$ in exactly one point $P$. The other $n$ lines through $P$ would then cover the remaining $n + 1$ points of $H$, each at most one of them, which is impossible.

Example: in the Fano plane, of order $2$, let $H$ be the four points off a fixed line $\ell$. Any other line meets $\ell$ in one point, so it has exactly $2$ points in $H$; hence no three points of $H$ are collinear, and $\abs{H} = 4 = 2 + 2$.

Non-example: $PG(2, 3)$ has no hyperoval. More generally, a plane of odd order $n$ has none. Suppose $S$ were a hyperoval, and take a point $P$ outside $S$. Every line through $P$ meets $S$ in $0$ or $2$ points, so these lines split $S$ into pairs. Then $\abs{S}$ is even, while $n + 2$ is odd.

For $n$ even, the tangents of an oval all pass through one point, its nucleus, and adding the nucleus gives a hyperoval.