Out of $8$ points, $4$ points on one branch of a hyperbola and $4$ on the other but no $5$ ever in convex position,is it true?

Question

Are any $8$ points, $4$ points on one branch of a hyperbola, $4$ points on the other branch of the same hyperbola always such that no $5$ points are in convex position (form a convex shape)

Answer 1

More is true: If we have four points on a hyperbola, three of them on the same branch, and one on the other branch, then these four points do not form a convex quadrangle.

Proof. Consider the hyperbola $xy=1$, and assume $P_i=(x_i,y_i)$ $\>(1\leq i\leq3)$ in the first quadrant, with $x_1<x_2<x_3$. The lines $g_{12}=P_1\vee P_2$ and $g_{23}=P_2\vee P_3$ do not intersect the third quadrant. They form a wedge with vertex $P_2$, containing the full third quadrant, hence the full second branch of the hyperbola, in its interior. It follows that $P_2$ is an interior point of the triangle $\triangle(P_4P_3P_1)$.