Curve is $y=x^2-4$.
Minimum distance from origin is along the normal passing through origin.
Consider a point on curve $\left(h, h^2-4\right)$.
Slope of tangent is $2 x=2 h$ at $\left(h, h^2-4\right)$
$\therefore$ Slope of normal is $-\frac{1}{2 h}$.
$\therefore$ Equation of normal is $\left(y-\left(h^2-4\right)=-\frac{1}{2 h}(x-h)\right)$
This passes through $(0,0)$.
$$
\begin{aligned}
& \therefore & -h^2+4 & =\frac{1}{2} \\
\Rightarrow & & h & = \pm \sqrt{\frac{7}{2}}
\end{aligned}
$$
$\therefore$ Points on the curve at minimum distance from origin can be $\left( \pm \sqrt{\frac{7}{2}},-\frac{1}{2}\right)$.
$$
\therefore \text { Minimum distance }=\sqrt{\frac{7}{2}+\frac{1}{4}}=\frac{\sqrt{15}}{2}
$$