The diagram shows the curve $y = a + 6\cos bx$ for $0 \le x \le \frac{2\pi}{3}$, where $a$ and $b$ are positive constants. The curve meets the x-axis at the point $\left(\frac{2\pi}{3},0\right)$. The curve has a maximum point at $(0,9)$ and a minimum point at $M$.

1. Use the maximum point $(0, 9)$ to find $a$.

Substitute $x = 0$ and $y = 9$ into $y = a + 6\cos bx$:
$9 = a + 6\cos(0)$
$9 = a + 6(1)$
$a = 3$

2. Use the x-intercept to form an equation for $b$.

Substitute $x = \frac{2\pi}{3}$ and $y = 0$ into $y = 3 + 6\cos bx$:
$0 = 3 + 6\cos\left(b \cdot \frac{2\pi}{3}\right)$
$6\cos\left(\frac{2\pi b}{3}\right) = -3$
$\cos\left(\frac{2\pi b}{3}\right) = -0.5$

3. Solve for $b$ using the condition that the minimum point $M$ lies between $x=0$ and $x=\frac{2\pi}{3}$.

The basic angle is $\cos^{-1}(0.5) = \frac{\pi}{3}$.
The solutions for $\frac{2\pi b}{3}$ are in the second and third quadrants:
$\frac{2\pi b}{3} = \pi - \frac{\pi}{3}, \pi + \frac{\pi}{3}, \dots$
$\frac{2\pi b}{3} = \frac{2\pi}{3}, \frac{4\pi}{3}, \dots$
If $\frac{2\pi b}{3} = \frac{2\pi}{3}$, then $b = 1$. The curve would be $y = 3 + 6\cos x$, which has its minimum at $x = \pi$. This is outside the interval $0 \le x \le \frac{2\pi}{3}$.
If $\frac{2\pi b}{3} = \frac{4\pi}{3}$, then $b = 2$. The curve is $y = 3 + 6\cos 2x$, which has its minimum when $2x = \pi \implies x = \frac{\pi}{2}$. Since $0 < \frac{\pi}{2} < \frac{2\pi}{3}$, this satisfies the condition that $M$ is between $0$ and $\frac{2\pi}{3}$.
Therefore, $b = 2$.