Mathematics

The key is recognizing that any point on a hyperbola must satisfy its equation. Since $(k,4)$ lies on the curve, you can substitute $x=k$ and $y=4$ directly into the hyperbola equation and solve for $k$. This transforms a geometry problem into a straightforward algebra problem—just plug in and isolate the unknown.

The key is recognizing that any point on a hyperbola must satisfy its equation. Since $(k,4)$ lies on the curve, you can substitute $x=k$ and $y=4$ directly into the hyperbola equation and solve for $k$. This transforms a geometry problem into a straightforward algebra problem—just plug in and isolate the unknown.