The coordinates of the point on the curve $x^2=4y$, which is at least distance from the line $y=x-4$ is

Given: Coordinates of the point $x^2=4y$ which is at the least distance of $y=x-4$, 

To Find: The coordinates of the line

Evaluate the slope of the line $y=x-4$, hence the slope is , $m=1$

Thus, the shortest distance to the curve $x^2=4y$ would be attained at the point where the tangent to the curve has a slope $m=1$

Differentiate the equation of curve with respect to $x$

$x^2=4y$

$2x=4\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{x}{2}$

Consider , the slope should be $1$ , therefore,

$\frac{dy}{dx}=1$

$\frac{x}{2}=1$

$x=2$

Hence, substituting the value of $x$ in the curve equation,

$y=\frac{x^2}{4}$

$y=1$

Hence, if the The coordinate of the point on the curve $x^2=4y$ which is at least distance from the line $y=x-4$ is $(2,1)$. 

Therefore, the correct answer is option $\left(1\right)$.