The minimum distance of a point on the curve $$y = x^2 - 4$$ from the origin is

The problem is to find the minimum distance from a point on the curve $$ y = x^2 - 4 $$ to the origin (0, 0).

The distance $$ d $$ from any point $$(x, y)$$ to the origin is given by $$ d = \sqrt{x^2 + y^2} $$. Since the point lies on the curve, substitute $$ y = x^2 - 4 $$ into the distance formula:

$$ d = \sqrt{x^2 + (x^2 - 4)^2} $$

Expand the expression inside the square root:

$$ d = \sqrt{x^2 + (x^4 - 8x^2 + 16)} $$

Combine like terms:

$$ d = \sqrt{x^4 - 7x^2 + 16} $$

Since the square root function is increasing, minimizing $$ d $$ is equivalent to minimizing $$ d^2 $$ (as $$ d^2 $$ will have its minimum at the same point). Let $$ f(x) = d^2 = x^4 - 7x^2 + 16 $$.

To find the minimum of $$ f(x) $$, take the derivative with respect to $$ x $$ and set it to zero. The derivative is:

$$ f'(x) = 4x^3 - 14x $$

Factor out $$ 2x $$:

$$ f'(x) = 2x(2x^2 - 7) $$

Set $$ f'(x) = 0 $$:

$$ 2x(2x^2 - 7) = 0 $$

This gives two equations:

$$ 2x = 0 \quad \text{or} \quad 2x^2 - 7 = 0 $$

Solving these:

$$ x = 0 \quad \text{or} \quad x^2 = \frac{7}{2} $$

So the critical points are $$ x = 0 $$, $$ x = \sqrt{\frac{7}{2}} $$, and $$ x = -\sqrt{\frac{7}{2}} $$.

Now evaluate $$ f(x) $$ at these points.

At $$ x = 0 $$:

$$ f(0) = (0)^4 - 7(0)^2 + 16 = 16 $$

At $$ x = \sqrt{\frac{7}{2}} $$ or $$ x = -\sqrt{\frac{7}{2}} $$ (since $$ x^2 $$ is the same):

$$ f\left(\sqrt{\frac{7}{2}}\right) = \left(\frac{7}{2}\right)^2 - 7 \cdot \frac{7}{2} + 16 $$

Calculate each term:

$$ \left(\frac{7}{2}\right)^2 = \frac{49}{4} $$

$$ -7 \cdot \frac{7}{2} = -\frac{49}{2} = -\frac{98}{4} $$

$$ +16 = \frac{64}{4} $$

Combine the fractions:

$$ f\left(\sqrt{\frac{7}{2}}\right) = \frac{49}{4} - \frac{98}{4} + \frac{64}{4} = \frac{49 - 98 + 64}{4} = \frac{15}{4} $$

Similarly, $$ f\left(-\sqrt{\frac{7}{2}}\right) = \frac{15}{4} $$.

Compare the values: $$ f(0) = 16 $$ and $$ f\left(\pm \sqrt{\frac{7}{2}}\right) = \frac{15}{4} = 3.75 $$. Since $$ 3.75 < 16 $$, the minimum occurs at $$ x = \pm \sqrt{\frac{7}{2}} $$.

The minimum distance $$ d $$ is the square root of the minimum $$ d^2 $$:

$$ d = \sqrt{\frac{15}{4}} = \frac{\sqrt{15}}{2} $$

Comparing with the options, $$ \frac{\sqrt{15}}{2} $$ units corresponds to Option A.

Hence, the correct answer is Option A.