1. Point on the curve

   The curve is $y=x^2-4.$ So any point on the curve can be written as $(x,a0x^2-4).$

2. Distance from origin

   The distance of the point $(x,a0x^2-4)$ from the origin $(0,0)$ is $d=\sqrt{x^2+(x^2-4)^2}.$

   To minimize $d$, it is easier to minimize $d^2$.

   Let $D=x^2+(x^2-4)^2.$

3. Expand $D$

   $D=x^2+x^4-8x+16 \quad \text{(incorrect expansion? let's do carefully)}$

   Since $(x^2-4)^2=x^4-8x^2+16,$ therefore $D=x^2+x^4-8x^2+16=x^4-7x^2+16.$

4. Differentiate

   $\frac{dD}{dx}=4x^3-14x=2x(2x^2-7).$

   Critical points are given by $2x(2x^2-7)=0$ so $x=0 \quad \text{or} \quad x^2=\frac{7}{2}.$

5. Check minimum value

   Evaluate $D$ at these critical points:

   • At $x=0$: $D=0^4-7(0)^2+16=16.$

   • At $x^2=\frac{7}{2}$:

     =\frac{49}{4}-\frac{49}{2}+16 =\frac{49-98+64}{4} =\frac{15}{4}.$$

   Since $\frac{15}{4}<16$, the minimum value of $D$ is $D_{\min}=\frac{15}{4}.$

6. Minimum distance

   Therefore, $d_{\min}=\sqrt{\frac{15}{4}}=\frac{\sqrt{15}}{2}.$

7. Match with options

   $\frac{\sqrt{15}}{2}$ corresponds to Option C.

Final Answer: $\boxed{\frac{\sqrt{15}}{2}}$