The area (in sq. units) of the largest rectangle $$ABCD$$ whose vertices $$A$$ and $$B$$ lie on the $$x$$-axis and vertices $$C$$ and $$D$$ lie on the parabola, $$y = x^2 - 1$$ below the $$x$$-axis, is:

1. Understand the Geometry of the Curve

The given parabola equation is:

$$y = x^2 - 1$$

This is a vertically opening parabola with its vertex at $$(0, -1)$$. It intersects the $$x$$-axis at $$x = 1$$ and $$x = -1$$. The region of the parabola that lies below the $$x$$-axis corresponds to the interval where $$-1 < x < 1$$, meaning $$y$$ is negative.

2. Set Up the Coordinates of the Rectangle

By symmetry about the $$y$$-axis, let the vertices of the rectangle be:

$$B = (x, 0)$$ on the positive $$x$$-axis, where $$0 < x < 1$$

$$A = (-x, 0)$$ on the negative $$x$$-axis

$$C = (x, x^2 - 1)$$ on the parabola in the fourth quadrant

$$D = (-x, x^2 - 1)$$ on the parabola in the third quadrant

3. Formulate the Area Function

The length of the rectangle along the $$x$$-axis is:

$$\text{Length} = x - (-x) = 2x$$

The height of the rectangle is the absolute distance from the $$x$$-axis down to the curve. Since $$y$$ is negative in this region, the height is:

$$\text{Height} = 0 - y = -(x^2 - 1) = 1 - x^2$$

The area $$A(x)$$ of the rectangle can be expressed as a function of $$x$$:

$$A(x) = \text{Length} \times \text{Height}$$

$$A(x) = 2x(1 - x^2)$$

$$A(x) = 2x - 2x^3$$

4. Maximize the Area Using Differentiation

To find the value of $$x$$ that yields the maximum area, differentiate $$A(x)$$ with respect to $$x$$ and set it to $$0$$:

$$A'(x) = \frac{d}{dx}(2x - 2x^3) = 2 - 6x^2$$

Setting the first derivative to zero:

$$2 - 6x^2 = 0$$

$$6x^2 = 2$$

$$x^2 = \frac{1}{3}$$

$$x = \frac{1}{\sqrt{3}}$$

Since the second derivative $$A''(x) = -12x$$ is strictly negative for $$x = \frac{1}{\sqrt{3}}$$, this point confirms a local maximum.

5. Compute the Maximum Area

Substitute $$x = \frac{1}{\sqrt{3}}$$ back into the area function:

$$A\left(\frac{1}{\sqrt{3}}\right) = 2\left(\frac{1}{\sqrt{3}}\right) - 2\left(\frac{1}{\sqrt{3}}\right)^3$$

$$A\left(\frac{1}{\sqrt{3}}\right) = \frac{2}{\sqrt{3}} - \frac{2}{3\sqrt{3}}$$

$$A\left(\frac{1}{\sqrt{3}}\right) = \frac{6 - 2}{3\sqrt{3}} = \frac{4}{3\sqrt{3}}$$