If $$P$$ is a point on the parabola $$y = x^2 + 4$$ which is closest to the straight line $$y = 4x - 1$$, then the co-ordinates of $$P$$ are:

We need to find the point $$P$$ on the parabola $$y = x^2 + 4$$ that is closest to the straight line $$y = 4x - 1$$.

The line $$y = 4x - 1$$ can be rewritten as $$4x - y - 1 = 0$$. The distance from any point $$(x_0, y_0)$$ on the parabola to this line is $$d = \frac{|4x_0 - y_0 - 1|}{\sqrt{16 + 1}} = \frac{|4x_0 - y_0 - 1|}{\sqrt{17}}$$.

Since $$y_0 = x_0^2 + 4$$, the distance becomes $$d = \frac{|4x_0 - x_0^2 - 4 - 1|}{\sqrt{17}} = \frac{|{-x_0^2 + 4x_0 - 5}|}{\sqrt{17}} = \frac{x_0^2 - 4x_0 + 5}{\sqrt{17}}$$, where we dropped the absolute value because the discriminant of $$-x_0^2 + 4x_0 - 5$$ is $$16 - 20 = -4 < 0$$, so $$x_0^2 - 4x_0 + 5 > 0$$ for all real $$x_0$$.

To minimize $$d$$, we minimize $$g(x_0) = x_0^2 - 4x_0 + 5$$. Differentiating, $$g'(x_0) = 2x_0 - 4 = 0$$, giving $$x_0 = 2$$. Since $$g''(x_0) = 2 > 0$$, this is indeed a minimum.

Substituting $$x_0 = 2$$ into the parabola equation, $$y_0 = (2)^2 + 4 = 4 + 4 = 8$$.

We can verify: the minimum distance is $$\frac{4 - 8 + 5}{\sqrt{17}} = \frac{1}{\sqrt{17}}$$. Also, the slope of the tangent to the parabola at $$x = 2$$ is $$\frac{dy}{dx}\bigg|_{x=2} = 2(2) = 4$$, which equals the slope of the given line, confirming that the tangent at the closest point is parallel to the line.

Therefore, the coordinates of $$P$$ are $$(2, 8)$$.