The curve C has parametric equations $x = 3t - 4, \, y = 5 - \frac{6}{t}, \ t > 0$ (a) Find $\frac{dy}{dx}$ in terms of t (b) The point P lies on C where $t = \frac{1}{2}$ - Edexcel - A-Level Maths Pure - Question 3 - 2017 - Paper 5

First, substitute $t = \frac{1}{2}$ into the parametric equations:

1. $x = 3\left(\frac{1}{2}\right) - 4 = -\frac{5}{2}$
2. $y = 5 - \frac{6}{\frac{1}{2}} = 5 - 12 = -7$

So the point P is $(-\frac{5}{2}, -7)$.

Next, find the slope of the tangent:
$\frac{dy}{dx}\bigg|_{t=\frac{1}{2}} = \frac{2}{\left(\frac{1}{2}\right)^2} = 8.$

Using the point-slope form of the line, the equation of the tangent is:
[y + 7 = 8\left(x + \frac{5}{2}\right).]
Solving this gives:
[y = 8x + 12 - 7 = 8x + 5.]
Thus, $p = 8$ and $q = 5$.

We start with the parametric equations:

1. From $x = 3t - 4$, we can express $t$ in terms of $x$:
   $t = \frac{x + 4}{3}.$
2. Substitute this into the $y$ equation:
   $y = 5 - \frac{6}{t} = 5 - \frac{6}{\frac{x + 4}{3}} = 5 - \frac{18}{x + 4}.$
3. Combine terms:
   $y = 5 - \frac{18}{x + 4} = \frac{5(x + 4)}{x + 4} - \frac{18}{x + 4} = \frac{5x + 20 - 18}{x + 4} = \frac{5x + 2}{x + 4}.$
   Thus, we have the form $y = \frac{ax + b}{x + 4}$ with $a = 5$ and $b = 2$.