Planar cone defined by a quadratic inequality

Let $n,k$ be nonzero orthogonal vectors in $\mathbb R^2$ , and let $a,b>0$ . The inequalities

n\cdot T>0,\qquad a(k\cdot T)^2<b(n\cdot T)^2

are equivalent to

n\cdot T>\sqrt{a/b}\,|k\cdot T|.

They define an open convex wedge invariant under positive scaling. Its closure is a convex cone containing zero.

Why the sign matters

The quadratic inequality without $n\cdot T>0$ also allows the opposite wedge. The positive longitudinal condition chooses one component. Replacing the strict inequalities by a uniform gap on $T/|T|$ supplies a directional margin that remains meaningful as $|T|$ tends to zero.