If P is a point on the parabola y = x ^ 2 + 4 which is closest to...

$P: y = {x}^{2} + 4$

$k = {h}^{2} + 4$

$L: y = 4x - 1$

$y - 4x + 1 = 0$

$D = AB = \left|\dfrac{k - 4h + 1}{\sqrt{17}}\right| = \left|\dfrac{{h}^{2} + 4 - 4h + 1}{\sqrt{17}}\right|$

$\dfrac{d\left(D\right)}{dh} = \dfrac{2h - 4}{\sqrt{17}} = 0$

$h = 2$

$\dfrac{{d}^{2}\left(D\right)}{d{h}^{2}} = \dfrac{2}{\sqrt{17}} > 0$

$∴ k = 4 + 4 = 8$

$∴$ Point \left(2, 8\right)