Given a line segment $ PQ$ moving on the parabola $ y \equal{} x^2$ with end points on the parabola. The area of the figure surrounded by $ PQ$ and the parabola is always equal to $ \frac {4}{3}$. Find the equation of the locus of the mid point $ M$ of $ PQ$.
1. Let $\alpha$ and $\beta$ ($\alpha < \beta$) be the $x$-coordinates of the points of intersection between the parabola $y = x^2$ and the line $PQ: y = l(x)$. We can write the equation of the parabola and the line as:
   \[
   x^2 - l(x) \equiv (x - \alpha)(x - \beta)
   \]
2. The area between the line segment $PQ$ and the parabola $y = x^2$ is given to be $\frac{4}{3}$. This area can be calculated using the integral:
   \[
   \text{Area} = \int_{\alpha}^{\beta} (x^2 - l(x)) \, dx
   \]
   Since $x^2 - l(x) = (x - \alpha)(x - \beta)$, we have:
   \[
   \text{Area} = \int_{\alpha}^{\beta} (x - \alpha)(x - \beta) \, dx
   \]
3. Evaluating the integral:
   \[
   \int_{\alpha}^{\beta} (x - \alpha)(x - \beta) \, dx = \int_{\alpha}^{\beta} (x^2 - (\alpha + \beta)x + \alpha \beta) \, dx
   \]
   \[
   = \left[ \frac{x^3}{3} - \frac{(\alpha + \beta)x^2}{2} + \alpha \beta x \right]_{\alpha}^{\beta}
   \]
   \[
   = \left( \frac{\beta^3}{3} - \frac{(\alpha + \beta)\beta^2}{2} + \alpha \beta \beta \right) - \left( \frac{\alpha^3}{3} - \frac{(\alpha + \beta)\alpha^2}{2} + \alpha \beta \alpha \right)
   \]
   Simplifying, we get:
   \[
   = \frac{\beta^3}{3} - \frac{\alpha^3}{3} - \frac{(\alpha + \beta)(\beta^2 - \alpha^2)}{2} + \alpha \beta (\beta - \alpha)
   \]
   \[
   = \frac{\beta^3 - \alpha^3}{3} - \frac{(\alpha + \beta)(\beta - \alpha)(\beta + \alpha)}{2} + \alpha \beta (\beta - \alpha)
   \]
   \[
   = \frac{(\beta - \alpha)(\beta^2 + \alpha \beta + \alpha^2)}{3} - \frac{(\beta - \alpha)(\alpha + \beta)^2}{2} + \alpha \beta (\beta - \alpha)
   \]
   \[
   = (\beta - \alpha) \left( \frac{\beta^2 + \alpha \beta + \alpha^2}{3} - \frac{(\alpha + \beta)^2}{2} + \alpha \beta \right)
   \]
   \[
   = (\beta - \alpha) \left( \frac{\beta^2 + \alpha \beta + \alpha^2}{3} - \frac{\alpha^2 + 2\alpha \beta + \beta^2}{2} + \alpha \beta \right)
   \]
   \[
   = (\beta - \alpha) \left( \frac{2\beta^2 + 2\alpha^2 + 6\alpha \beta - 3\alpha^2 - 6\alpha \beta - 3\beta^2}{6} \right)
   \]
   \[
   = (\beta - \alpha) \left( \frac{-\beta^2 - \alpha^2}{6} \right)
   \]
   \[
   = \frac{(\beta - \alpha)^3}{6}
   \]
   Given that the area is $\frac{4}{3}$, we have:
   \[
   \frac{(\beta - \alpha)^3}{6} = \frac{4}{3}
   \]
   \[
   (\beta - \alpha)^3 = 8
   \]
   \[
   \beta - \alpha = 2
   \]
4. Let $M(X, Y)$ be the midpoint of $PQ$. Since $P(\alpha, \alpha^2)$ and $Q(\beta, \beta^2)$, we have:
   \[
   X = \frac{\alpha + \beta}{2}, \quad Y = \frac{\alpha^2 + \beta^2}{2}
   \]
   Therefore:
   \[
   \alpha + \beta = 2X, \quad \alpha^2 + \beta^2 = 2Y
   \]
5. Using the identity $(\alpha + \beta)^2 + (\alpha - \beta)^2 = 2(\alpha^2 + \beta^2)$, we get:
   \[
   (2X)^2 + 2^2 = 2(2Y)
   \]
   \[
   4X^2 + 4 = 4Y
   \]
   \[
   Y = X^2 + 1
   \]
   Thus, the equation of the locus of the midpoint $M$ is:
   \[
   y = x^2 + 1
   \]

The final answer is $\boxed{y = x^2 + 1}$