Length of loop

From the graph, we note that the loop is between $3 \le x \le 6$ . The length of the loop is given by:

$\begin{aligned} s & = 2 \int_3^6 \sqrt{1+\color{#3D99F6} {\left(\frac{dy}{dx}\right)^2}} dx \\ & = 2 \int_3^6 \sqrt{1+\color{#3D99F6} {\frac{(x-4)^2}{4(x-3)}}} dx \quad \quad \color{#3D99F6}{[\text{See Note}]} \\ & = 2 \int_3^6 \sqrt{\frac{4x-12 + x^2 - 8x + 16}{4(x-3)}} dx = 2 \int_3^6 \sqrt{\frac{x^2 - 4x + 4}{4(x-3)}} dx \\ & = \int_3^6 \frac{x - 2}{\sqrt{x-3}} dx \quad \quad \color{#D61F06} {[\text{Let }u^2 = x - 3 \Rightarrow x = u^2 + 3 \Rightarrow dx = 2u\space du]} \\ & = \int _{\color{#D61F06}{0}}^{\color{#D61F06}{\sqrt{3}}} \frac{\color{#D61F06}{u^2+3}- 2}{\sqrt{\color{#D61F06}{u^2}}} \color{#D61F06}{(2u)du} = 2 \int_0^{\sqrt{3}} (u^2+1) du \\ & = 2 \left[\frac{u^3}{3} + u \right]_0^{\sqrt{3}} = 2 \left[\sqrt{3}+\sqrt{3}\right] = 4\sqrt{3} \end{aligned}$

$\Rightarrow a \times b = 4 \times 3 = \boxed{12}$

$$

$\color{#3D99F6} {\text{Note:}}$

$\begin{aligned} 9y^2 & = (x-3)(x-6)^2 \\ 18 y \frac{dy}{dx} & = (x-6)^2 + (x-3)(2)(x-6) \\ \frac{dy}{dx} & = \frac{(x-6)(x-6+2x-6)}{18 y} = \frac{(x-6)(x-4)}{6y} \\ \color{#3D99F6} {\left( \frac{dy}{dx} \right)^2} & = \frac{(x-6)^2(x-4)^2}{36y^2} = \frac{(x-6)^2(x-4)^2}{4(x-3)(x-6)^2} = \color{#3D99F6}{\frac{(x-4)^2}{4(x-3)}} \end{aligned}$