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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:
(1) $(-2, 8)$
(2) $(1, 5)$
(3) $(3, 13)$
(4) $(2, 8)$
$\frac{dy}{dx}\bigg|_P = 4 \Rightarrow 2x_1 = 4 \Rightarrow x_1 = 2$. Point is $(2, 8)$.
BETA
