A square hole of depth $h$ whose base is of length $a$ is given. A dog is tied to the center of the square at the bottom of the hole by a rope of length $L >\sqrt{2a^2+h^2}$, and walks on the ground around the hole. The edges of the hole are smooth, so that the rope can freely slide along it. Find the shape and area of the territory accessible to the dog (whose size is neglected).
1. **Define the problem and setup the coordinates:**
   - Let \(ABCD\) be the upper square of the hole with side length \(a\).
   - Let \(O\) be the center of the bottom square.
   - Let \(T\) be a variable point on the edge \(AB\) through which the string passes.
   - Let \(P\) be the point where the dog is standing, such that the string remains taut.
   - Let \(O'\) be the center of the top square, which coincides with \(O\) when viewed from the top.

2. **Determine the length of the string inside the hole:**
   - The length of the string inside the hole can be calculated using the Pythagorean theorem. The distance from \(O\) to the edge of the hole is \(\frac{a}{2}\), and the depth of the hole is \(h\).
   - Therefore, the length of the string inside the hole is:
     \[
     \sqrt{\left(\frac{a}{2}\right)^2 + h^2} = \frac{1}{2}\sqrt{a^2 + 4h^2}
     \]

3. **Determine the length of the string on the top plane:**
   - Let the total length of the string be \(L\). The length of the string on the top plane is:
     \[
     \ell = L - \frac{1}{2}\sqrt{a^2 + 4h^2}
     \]

4. **Find the locus of point \(P\):**
   - Let \(O'\) be the origin \((0,0)\) and the axes be parallel to the sides of the square.
   - Let \(P \equiv (h, k)\).
   - The point \(T\) where the string touches the edge \(AB\) can be found by the intersection of the line \(PO'\) with \(AB\). The line \(PO'\) has the equation \(y = \frac{k}{h}x\).
   - The edge \(AB\) has the equation \(x = \frac{a}{2}\).
   - Therefore, the coordinates of \(T\) are:
     \[
     T \equiv \left(\frac{a}{2}, \frac{ka}{2h}\right)
     \]

5. **Calculate the distance \(PT\):**
   - The distance \(PT\) is given by:
     \[
     PT^2 = \left(h - \frac{a}{2}\right)^2 + \left(k - \frac{ka}{2h}\right)^2
     \]
   - Simplifying the expression:
     \[
     PT^2 = \left(h - \frac{a}{2}\right)^2 + \left(k - \frac{ka}{2h}\right)^2 = \ell^2
     \]

6. **Form the equation of the locus:**
   - Substituting \(a = 2b\) into the equation:
     \[
     (x - b)^2 + \frac{y^2(x - b)^2}{x^2} = \ell^2
     \]

7. **Reflect the right branch of the plot about the axes:**
   - The final shape of the territory accessible to the dog is obtained by reflecting the right branch of the plot about the axes.

The final answer is \( \boxed{ (x - b)^2 + \frac{y^2(x - b)^2}{x^2} = \ell^2 } \)