JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Application of Derivatives

For the point on the parabola $y=x^2+4$ closest to the line $y=4x-1$, the tangent to the parabola at that point must be parallel to the given line.

The derivative of $y=x^2+4$ is $\frac{dy}{dx}=2x$.

For the tangent to be parallel to $y=4x-1$ (slope = 4), we need:
$2x=4\Rightarrow x=2$

When $x=2$: $y=2^2+4=8$

Therefore, the point $P$ is $(2,8)$.