The diagram shows part of the curve $y = 27 + \frac{8}{(x-2)^3}$ which crosses the axes at points $A$ and $B$.

1. Find the first derivative of the curve equation

$y = 27 + 8(x-2)^{-3}$
$\frac{dy}{dx} = 8(-3)(x-2)^{-4} = -\frac{24}{(x-2)^4}$

2. State the condition for a stationary point and explain why it is not met

For a stationary point, $\frac{dy}{dx} = 0$.
However, $-\frac{24}{(x-2)^4} = 0$ has no solution because the numerator is $-24 \neq 0$.
Therefore, the curve does not have a stationary point.

1. Find the coordinates of $A$ by setting $y = 0$

At $A$, $y = 0$:
$0 = 27 + \frac{8}{(x-2)^3}$
$\frac{8}{(x-2)^3} = -27$
$(x-2)^3 = -\frac{8}{27}$
$x - 2 = -\frac{2}{3}$
$x = \frac{4}{3}$
Coordinates of $A$ are $\left(\frac{4}{3}, 0\right)$.

2. Find the coordinates of $B$ by setting $x = 0$

At $B$, $x = 0$:
$y = 27 + \frac{8}{(0-2)^3}$
$y = 27 + \frac{8}{-8}$
$y = 27 - 1 = 26$
Coordinates of $B$ are $(0, 26)$.

1. Find the coordinates of point $B$ (the $y$-intercept).

At $B$, $x = 0$.
$y = 27 + \frac{8}{(0-2)^3} = 27 + \frac{8}{-8} = 26$
So $B$ is $(0, 26)$.

2. Find the derivative $\frac{dy}{dx}$ and evaluate it at $x = 0$ to find the gradient of the tangent.

$y = 27 + 8(x-2)^{-3}$
$\frac{dy}{dx} = -24(x-2)^{-4} = -\frac{24}{(x-2)^4}$
At $x = 0$, gradient of tangent $m_t = -\frac{24}{(-2)^4} = -\frac{24}{16} = -\frac{3}{2}$

3. Find the gradient of the normal and the equation of the normal.

Gradient of normal $m_n = -\frac{1}{m_t} = \frac{2}{3}$
Equation of normal at $B(0, 26)$:
$y - 26 = \frac{2}{3}(x - 0)$
$y = \frac{2}{3}x + 26$

1. Find the coordinates of $C$ using the equation of the normal from part (c)

From part (c), the equation of the normal at $B(0, 26)$ is $y = \frac{2}{3}x + 26$.
At $C$, the normal cuts the $x$-axis, so $y = 0$.
$0 = \frac{2}{3}x + 26$
$\frac{2}{3}x = -26$
$x = -39$
So, $C$ is $(-39, 0)$.

2. Calculate the area of triangle $BOC$

The vertices of triangle $BOC$ are $B(0, 26)$, $O(0, 0)$, and $C(-39, 0)$.
The base of the triangle along the $x$-axis is $OC = 39$ units.
The perpendicular height along the $y$-axis is $OB = 26$ units.
$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$
$\text{Area} = \frac{1}{2} \times 39 \times 26$
$\text{Area} = 507$