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The smallest possible distance is 5*sqrt(5).

y = 1/2x^2 - 4

Suppose that $(x_0, y_0)$ is a point on this graph closest to the origin.

Hence:

$y_0 = \frac12(x_0)^2 - 4.$

Using this information, we need to find 

$\sqrt{x_0^2 + y_0^2}$

We can plug in $x_0^2$ with $2y_0 + 8$ (Make sure you see how I got this)

Hence, the distance between the points will be:

$\sqrt(y_0^2 +2y_0 + 8)$

Since this distance is to be minimized by how we defined the point (x_0, y_0), we can find the vertex of the that quadratic in y_0. If you complete the square, you will recieve that the parabola reaches its minimum when $y_0 = -1$.

Can you solve the problem from here?