The curve C has parametric equations $x = 3t - 4, \, y = 5 - \frac{6}{t} \, (t > 0)$ - Edexcel - A-Level Maths Pure - Question 3 - 2017 - Paper 5

To find the coordinates of point P when $t = \frac{1}{2}$:

1. Substitute $t = \frac{1}{2}$ into the equations: $x = 3(\frac{1}{2}) - 4 = \frac{3}{2} - 4 = -\frac{5}{2}$ $y = 5 - \frac{6}{\frac{1}{2}} = 5 - 12 = -7$

Thus, point P is at $\left(-\frac{5}{2}, -7\right)$.

The slope of the tangent line can be evaluated using our previous calculation:

1. Substitute $t = \frac{1}{2}$ into $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{2}{(\frac{1}{2})^2} = \frac{2}{\frac{1}{4}} = 8$

Now, using the point-slope form of the line, $y - y_1 = m(x - x_1)$:

1. Substitute $m = 8$, $x_1 = -\frac{5}{2}$, and $y_1 = -7$: $y + 7 = 8\left(x + \frac{5}{2}\right)$ $y = 8x + 20 - 7$ $y = 8x + 13$

Thus, $p = 8$ and $q = 13$.

To derive the Cartesian equation:

1. Start from the parametrized equations: $x = 3t - 4 \Rightarrow t = \frac{x + 4}{3}$

2. Substitute into the equation for y: $y = 5 - \frac{6}{t} = 5 - \frac{6}{\frac{x + 4}{3}} = 5 - \frac{18}{x + 4}$

3. Combine terms: $y = \frac{5(x + 4) - 18}{x + 4} = \frac{5x + 20 - 18}{x + 4} = \frac{5x + 2}{x + 4}$

Thus, in the form $y = \frac{ax + b}{x + 4}$, we find $a = 5$ and $b = 2$.