The diagram shows part of the curve $y = \frac{1}{\left(\frac{3}{4}x - 5\right)^2} + 1$. $P$ is the point on the curve where $x = 6$. The tangent to the curve at the point $P$ meets the $x$-axis at $Q$.

1. Find the coordinates of $P$ and the gradient of the tangent at $P$.

When $x = 6$,
$y = \frac{1}{\left(\frac{3}{4}(6) - 5\right)^2} + 1 = \frac{1}{(-0.5)^2} + 1 = 5$
So $P$ is $(6, 5)$.

$\frac{dy}{dx} = -2\left(\frac{3}{4}x - 5\right)^{-3} \left(\frac{3}{4}\right) = -\frac{3}{2}\left(\frac{3}{4}x - 5\right)^{-3}$
At $x = 6$, $\frac{dy}{dx} = -\frac{3}{2}(-0.5)^{-3} = 12$

2. Find the coordinates of $Q$ (from part a).

Equation of tangent at $P$:
$y - 5 = 12(x - 6)$
$y = 12x - 67$
At $Q$, $y = 0 \implies 12x = 67 \implies x = \frac{67}{12}$
So $Q$ is $\left(\frac{67}{12}, 0\right)$.

3. Calculate the area under the curve from $x = 0$ to $x = 6$.

Area under curve $= \int_0^6 \left( \left(\frac{3}{4}x - 5\right)^{-2} + 1 \right) dx$
$= \left[ \frac{\left(\frac{3}{4}x - 5\right)^{-1}}{-1 \cdot \frac{3}{4}} + x \right]_0^6$
$= \left[ -\frac{4}{3\left(\frac{3}{4}x - 5\right)} + x \right]_0^6$
$= \left( -\frac{4}{3(-0.5)} + 6 \right) - \left( -\frac{4}{3(-5)} + 0 \right)$
$= \left( \frac{8}{3} + 6 \right) - \left( \frac{4}{15} \right)$
$= \frac{26}{3} - \frac{4}{15} = \frac{126}{15} = \frac{42}{5}$

4. Calculate the area of the triangle under the tangent and subtract it to find the shaded area.

The area under the tangent from $x = \frac{67}{12}$ to $x = 6$ forms a right-angled triangle.
Base of triangle $= 6 - \frac{67}{12} = \frac{5}{12}$
Height of triangle $= 5$
Area of triangle $= \frac{1}{2} \times \frac{5}{12} \times 5 = \frac{25}{24}$

Shaded Area $= \text{Area under curve} - \text{Area of triangle}$
$= \frac{42}{5} - \frac{25}{24} = \frac{1008 - 125}{120} = \frac{883}{120}$