Let $C$ be the point of intersection of the tangent lines $l,\ m$ at $A(a,\ a^2),\ B(b,\　b^2)\ (a<b)$ on the parabola $y=x^2$ respectively.
When $C$ moves on the parabola $y=\frac 12x^2-x-2$, find the minimum area bounded by 2 lines $l,\ m$ and the parabola $y=x^2$.
1. **Find the equations of the tangent lines at points \(A(a, a^2)\) and \(B(b, b^2)\) on the parabola \(y = x^2\):**
   - The slope of the tangent line at \(A(a, a^2)\) is given by the derivative of \(y = x^2\) at \(x = a\), which is \(2a\). Therefore, the equation of the tangent line at \(A\) is:
     \[
     y - a^2 = 2a(x - a) \implies y = 2ax - a^2
     \]
   - Similarly, the slope of the tangent line at \(B(b, b^2)\) is \(2b\). Therefore, the equation of the tangent line at \(B\) is:
     \[
     y - b^2 = 2b(x - b) \implies y = 2bx - b^2
     \]

2. **Find the coordinates of the intersection point \(C\) of the tangent lines \(l\) and \(m\):**
   - Set the equations of the tangent lines equal to each other to find the \(x\)-coordinate of \(C\):
     \[
     2ax - a^2 = 2bx - b^2 \implies 2ax - 2bx = a^2 - b^2 \implies x = \frac{a^2 - b^2}{2(a - b)} = \frac{a + b}{2}
     \]
   - Substitute \(x = \frac{a + b}{2}\) back into either tangent line equation to find the \(y\)-coordinate of \(C\):
     \[
     y = 2a\left(\frac{a + b}{2}\right) - a^2 = a(a + b) - a^2 = ab
     \]
   - Therefore, the coordinates of \(C\) are:
     \[
     C\left(\frac{a + b}{2}, ab\right)
     \]

3. **Verify that \(C\) lies on the parabola \(y = \frac{1}{2}x^2 - x - 2\):**
   - Substitute \(x = \frac{a + b}{2}\) and \(y = ab\) into the equation of the parabola:
     \[
     ab = \frac{1}{2}\left(\frac{a + b}{2}\right)^2 - \frac{a + b}{2} - 2
     \]
   - Simplify the equation:
     \[
     ab = \frac{1}{2} \cdot \frac{(a + b)^2}{4} - \frac{a + b}{2} - 2 = \frac{(a + b)^2}{8} - \frac{a + b}{2} - 2
     \]
   - This confirms that \(C\) lies on the given parabola.

4. **Calculate the area bounded by the tangent lines \(l\), \(m\), and the parabola \(y = x^2\):**
   - The area \(S\) is given by the sum of the integrals of the difference between the parabola \(y = x^2\) and the tangent lines from \(a\) to \(\frac{a + b}{2}\) and from \(\frac{a + b}{2}\) to \(b\):
     \[
     S = \left| \int_{a}^{\frac{a + b}{2}} (x^2 - (2ax - a^2)) \, dx \right| + \left| \int_{\frac{a + b}{2}}^{b} (x^2 - (2bx - b^2)) \, dx \right|
     \]
   - Simplify the integrals:
     \[
     \int_{a}^{\frac{a + b}{2}} (x^2 - 2ax + a^2) \, dx = \int_{a}^{\frac{a + b}{2}} (x - a)^2 \, dx
     \]
     \[
     \int_{\frac{a + b}{2}}^{b} (x^2 - 2bx + b^2) \, dx = \int_{\frac{a + b}{2}}^{b} (x - b)^2 \, dx
     \]
   - Evaluate the integrals:
     \[
     \int (x - a)^2 \, dx = \frac{(x - a)^3}{3}
     \]
     \[
     \left. \frac{(x - a)^3}{3} \right|_{a}^{\frac{a + b}{2}} = \frac{\left(\frac{a + b}{2} - a\right)^3}{3} = \frac{\left(\frac{b - a}{2}\right)^3}{3} = \frac{(b - a)^3}{24}
     \]
     \[
     \int (x - b)^2 \, dx = \frac{(x - b)^3}{3}
     \]
     \[
     \left. \frac{(x - b)^3}{3} \right|_{\frac{a + b}{2}}^{b} = \frac{\left(b - \frac{a + b}{2}\right)^3}{3} = \frac{\left(\frac{b - a}{2}\right)^3}{3} = \frac{(b - a)^3}{24}
     \]
   - Therefore, the total area \(S\) is:
     \[
     S = \frac{(b - a)^3}{24} + \frac{(b - a)^3}{24} = \frac{(b - a)^3}{12}
     \]

The final answer is \(\boxed{\frac{(b - a)^3}{12}}\).