Answer: $S = \frac{(m+n)^2}{\sqrt{m-n}}$.

Let $x_R, y_R$ denote the coordinates of point $R$. Let $y = kx + a$ and $y = kx + b$ be the equations of the lines $AD$ and $BC$, respectively. Then
$$\begin{cases} x_A^2 = y_A = kx_A + a, \\ x_D^2 = y_D = kx_D + a, \end{cases} \begin{cases} x_B^2 = y_B = kx_B + b, \\ x_C^2 = y_C = kx_C + b. \end{cases}$$
So,

$$k = x_A + x_D = x_B + x_C, \quad (1)$$
$$a = -x_A x_D, \quad b = -x_B x_C. \quad (2)$$
Since $K$ and $L$ are the midpoints of the sides $AD$ and $BC$ respectively, we see that
$$x_K = 0.5(x_A + x_D), \quad x_L = 0.5(x_B + x_C). \quad (3)$$
It is well-known fact that point $M$ of intersection of the diagonals of the trapezoid $ABCD$ lies on the segment joining the midpoints of the trapezoid bases. Taking into account (3) we see that the segment $KL$ containing $M$ is perpendicular to the axis $Ox$, and $x_M = x_K = x_L$. Let $\alpha$ be the angle between the lines $AD$, $BC$ and the positive direction of $Ox$, then $\tan \alpha = k$. If $LN$ is an altitude of the trapezoid $ABCD$, then $\angle NLK = \alpha$ ($LN \perp AD, KL \perp Ox$). Therefore $LN = KL \cos \alpha$. On the other hand, $AD = (y_D - y_A)/\cos \alpha$, $BC = (y_C - y_B)/\cos \alpha$. Therefore, the required area is equal to
$$\begin{aligned} S &= \frac{1}{2}(AD + BC)LN = \frac{1}{2}(y_D - y_A + y_C - y_B)KL \cdot \frac{\cos \alpha}{\sin \alpha} = [KL = m+n] = \\ &= \frac{1}{2}(y_D - y_A + y_C - y_B) \cdot \frac{m+n}{k} = \frac{1}{2} \left( \frac{y_D - y_A}{k} + \frac{y_C - y_B}{k} \right) (m+n) \stackrel{(1)}{=} \\ &= \frac{1}{2}(x_D - x_A + x_C - x_B)(m+n) \end{aligned} \quad (4)$$
Let $y = k_1x + c$ be the equation of the line $BD$. Then
$$\left\{ \begin{array}{l} x_B^2 = y_B = k_1 x_B + c, \\ x_D^2 = y_D = k_1 x_D + c, \end{array} \right. \quad \Rightarrow \quad k_1 = x_B + x_D, \quad c = -x_B x_D. \quad (5)$$
So, $y_M = 0.5k_1(x_A + x_D) + c = 0.5k_1(x_B + x_C) + c$. Therefore, from (1) - (3), and (5) it follows that
$$\begin{aligned} m &= y_K - y_M = \frac{1}{2}(x_A+x_D)(x_A+x_D) - x_A x_D - \frac{1}{2}(x_B+x_D)(x_A+x_D) + x_B x_D = \\ &= \frac{1}{2}(x_A + x_D)(x_A - x_B) - x_D(x_A - x_B) = \frac{1}{2}(x_A - x_B)(x_A - x_D). \end{aligned}$$
$$\begin{aligned} n &= y_M - y_L \stackrel{(1)}{=} \frac{1}{2}(x_B+x_D)(x_A+x_D) - x_B x_D - \frac{1}{2}(x_A+x_D)(x_A+x_D) + x_B x_C = \\ &= \frac{1}{2}(x_A + x_D)(x_B - x_A) - x_B(x_D - x_C) = [x_D - x_C \stackrel{(1)}{=} x_B - x_A] = \\ &= \frac{1}{2}(x_B - x_A)(x_A + x_D - 2x_B) \stackrel{(1)}{=} \frac{1}{2}(x_B - x_A)(x_C - x_B). \end{aligned}$$
Then $m+n = \frac{1}{2}(x_B - x_A)(x_C - x_B + x_D - x_A)$. Hence, (4) can be presented in the form
$$S = \frac{1}{2} \cdot \frac{2(m+n)}{x_B - x_A} (m+n) = \frac{(m+n)^2}{x_B - x_A}.$$
Note that
$$m-n = \frac{1}{2}(x_B - x_A)(x_D - x_A - x_C + x_B) = [x_D - x_C \stackrel{(1)}{=} x_B - x_A] = (x_B - x_A)^2.$$

Therefore, finally, we have $S = \frac{(m+n)^2}{x_B - x_A} = \frac{(m+n)^2}{\sqrt{m-n}}$.