Mathematics

The key is recognizing that any point on a parabola must satisfy its equation. Since $(k, 2\sqrt{6})$ lies on $y^2 = 24x$, you can substitute the $y$-coordinate into the equation and solve for $k$. This turns a geometric problem into a straightforward algebraic one—just plug in what you know and isolate what you don't.

The key is recognizing that any point on a parabola must satisfy its equation. Since $(k, 2\sqrt{6})$ lies on $y^2 = 24x$, you can substitute the $y$-coordinate into the equation and solve for $k$. This turns a geometric problem into a straightforward algebraic one—just plug in what you know and isolate what you don't.