Let $m$ be a real number. On a plane with the coordinate system, in which the origin is denoted by O, consider the parabola $y = x^2$ and the two points on it,

$$\mathrm{A}(a,\, ma+1), \quad \mathrm{B}(b,\, mb+1) \quad (a < 0 < b)$$

(1) The $x$-coordinates $a$ and $b$ of the two points A and B can be expressed in terms of $m$ as

$$a = \frac{m - \sqrt{D}}{\mathbf{A}}, \quad b = \frac{m + \sqrt{D}}{\mathbf{B}},$$

where the expression $D$ is

$$D = m^2 + \mathbf{C}.$$

(2) Let the coordinates of the point of intersection of the segment AB and the $y$-axis be denoted by $(0, c)$. Then $c = \mathbf{D}$.

(3) Further, when the area $S$ of the triangle OAB with the three vertices O, A and B is expressed in terms of $a$ and $b$, we have

$$S = \frac{1}{2}\mathbf{E},$$

where $E$ is the appropriate choice from among (0) $\sim$ (5).\\
(0) $a + b$\\
(1) $a - b$\\
(2) $b - a$\\
(3) $a^2 + b^2$\\
(4) $a^2 - b^2$\\
(5) $b^2 - a^2$

Also, when $S$ is represented in terms of $m$, we have

$$S = \frac{\mathbf{F}}{\mathbf{G}} \sqrt{m^2 + \mathbf{H}}.$$

Hence the value of $S$ is minimalized when $m = \mathbf{I}$, and its minimum value is $S = \mathbf{J}$.