The point $M$ varies on the segment $AB$ that measures $2$ m.
a) Find the equation and the graphical representation of the locus of the points of the plane whose coordinates, $x$, and $y$, are, respectively, the areas of the squares of sides $AM$ and $MB$ .
b) Find out what kind of curve it is. (Suggestion: make a $45^o$ axis rotation).
c) Find the area of the enclosure between the curve obtained and the coordinate axes.
### Part (a)
1. **Define the points and variables:**
   - Let $A(0,0)$ and $B(2,0)$ be the endpoints of the segment.
   - Let $M(\lambda, 0)$ be a point on the segment $AB$ where $0 \leq \lambda \leq 2$.
   - The coordinates $(x, y)$ represent the areas of the squares with sides $AM$ and $MB$ respectively.

2. **Express $x$ and $y$ in terms of $\lambda$:**
   - The length of $AM$ is $\lambda$, so the area of the square with side $AM$ is $x = \lambda^2$.
   - The length of $MB$ is $2 - \lambda$, so the area of the square with side $MB$ is $y = (2 - \lambda)^2$.

3. **Eliminate the parameter $\lambda$:**
   - From $x = \lambda^2$, we get $\lambda = \sqrt{x}$.
   - Substitute $\lambda = \sqrt{x}$ into $y = (2 - \lambda)^2$:
     \[
     y = (2 - \sqrt{x})^2 = 4 - 4\sqrt{x} + x
     \]

4. **Simplify the equation:**
   - Rearrange the equation to get:
     \[
     y = x + 4 - 4\sqrt{x}
     \]

### Part (b)
1. **Identify the type of curve:**
   - To identify the type of curve, we perform a $45^\circ$ axis rotation. Let $u$ and $v$ be the new coordinates after rotation:
     \[
     u = \frac{x + y}{\sqrt{2}}, \quad v = \frac{y - x}{\sqrt{2}}
     \]

2. **Substitute $x$ and $y$ into the new coordinates:**
   - From $y = x + 4 - 4\sqrt{x}$, we get:
     \[
     u = \frac{x + (x + 4 - 4\sqrt{x})}{\sqrt{2}} = \frac{2x + 4 - 4\sqrt{x}}{\sqrt{2}}
     \]
     \[
     v = \frac{(x + 4 - 4\sqrt{x}) - x}{\sqrt{2}} = \frac{4 - 4\sqrt{x}}{\sqrt{2}}
     \]

3. **Simplify the expressions:**
   - Let $t = \sqrt{x}$, then $x = t^2$ and the equations become:
     \[
     u = \frac{2t^2 + 4 - 4t}{\sqrt{2}}
     \]
     \[
     v = \frac{4 - 4t}{\sqrt{2}}
     \]

4. **Identify the curve:**
   - The resulting equations suggest that the curve is a parabola.

### Part (c)
1. **Find the area enclosed by the curve and the coordinate axes:**
   - The curve $y = x + 4 - 4\sqrt{x}$ intersects the $x$-axis at $x = 0$ and $x = 4$.
   - The area under the curve from $x = 0$ to $x = 4$ is given by:
     \[
     \int_{0}^{4} (x + 4 - 4\sqrt{x}) \, dx
     \]

2. **Evaluate the integral:**
   - Split the integral into three parts:
     \[
     \int_{0}^{4} x \, dx + \int_{0}^{4} 4 \, dx - \int_{0}^{4} 4\sqrt{x} \, dx
     \]
   - Compute each part:
     \[
     \int_{0}^{4} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{4} = \frac{16}{2} = 8
     \]
     \[
     \int_{0}^{4} 4 \, dx = 4x \bigg|_{0}^{4} = 4 \times 4 = 16
     \]
     \[
     \int_{0}^{4} 4\sqrt{x} \, dx = 4 \int_{0}^{4} x^{1/2} \, dx = 4 \left[ \frac{2}{3} x^{3/2} \right]_{0}^{4} = 4 \times \frac{2}{3} \times 8 = \frac{64}{3}
     \]

3. **Combine the results:**
   - The total area is:
     \[
     8 + 16 - \frac{64}{3} = 24 - \frac{64}{3} = \frac{72}{3} - \frac{64}{3} = \frac{8}{3}
     \]

The final answer is $\boxed{\frac{8}{3}}$