The curve $C_1$ has equation $y = \dfrac{ax^2+bx+3}{x+2}$, where $a$ and $b$ are constants. It is given that $C_1$ has an asymptote $y = x - 3$.

1. Rewrite the equation of the curve to isolate the linear part representing the oblique asymptote.

By algebraic manipulation or polynomial long division:
$\begin{aligned} y &= \frac{ax^2+bx+3}{x+2} \\ &= \frac{ax(x+2) - 2ax + bx + 3}{x+2} \\ &= \frac{ax(x+2) + (b-2a)x + 3}{x+2} \\ &= \frac{ax(x+2) + (b-2a)(x+2) - 2(b-2a) + 3}{x+2} \\ &= ax + (b-2a) + \frac{3-2b+4a}{x+2} \end{aligned}$

2. Identify the theoretical equation of the oblique asymptote.

As $x \to \pm\infty$, the fractional term $\frac{3-2b+4a}{x+2} \to 0$.
Therefore, the equation of the oblique asymptote is given by the linear portion:
$y = ax + (b-2a)$

3. Compare with the given asymptote to state the value of $a$ and show $b = -1$.

We are given that the asymptote is $y = x - 3$. Comparing the coefficients with $y = ax + (b-2a)$:

Coefficient of $x$:
$a = 1$

Constant term:
$b - 2a = -3$
Substitute $a = 1$ into the constant term equation:
$b - 2(1) = -3$
$b - 2 = -3$
$b = -1 \text{ (shown)}$

1. Substitute the constants from part (a) and rearrange the given equation into a quadratic equation in terms of $x$.

Using $a = 1$ and $b = -1$ from part (a), the equation of the curve is:
$y = \frac{x^2 - x + 3}{x + 2}$
Rearranging to form a quadratic equation in $x$:
$y(x + 2) = x^2 - x + 3$
$xy + 2y = x^2 - x + 3$
$x^2 - (y + 1)x + (3 - 2y) = 0$

2. Apply the discriminant condition for values that $y$ cannot take (i.e. no real roots for $x$).

For $y$ to be a value that the curve cannot take, the quadratic equation must have no real solutions for $x$. Therefore, its discriminant must be strictly less than $0$:
$\Delta < 0$
$(-(y + 1))^2 - 4(1)(3 - 2y) < 0$

3. Expand, factorise and solve the quadratic inequality.

Expanding and simplifying the inequality:
$(y^2 + 2y + 1) - 12 + 8y < 0$
$y^2 + 10y - 11 < 0$
Factorising the quadratic expression:
$(y + 11)(y - 1) < 0$
Solving this yields the set of values $y$ cannot take:
$-11 < y < 1$

1. Express the $y$-coordinate of the midpoint in terms of $x$.

Since the locus is traced as a single parameter $x$ varies, the $x$-coordinates of $P$ and $Q$ must both be $x$, giving $x_1 = x_2 = x$.

Let the midpoint of $P$ and $Q$ be $(X, Y)$. The $x$-coordinate of the midpoint is $X = x$. The $y$-coordinate is given by:
$Y = \frac{y_1 + y_2}{2}$
$Y = \frac{1}{2} \left( \frac{x^2-x+3}{x+2} + \frac{3x-x^2}{x+2} \right)$
$Y = \frac{1}{2} \left( \frac{x^2 - x + 3 + 3x - x^2}{x+2} \right)$
$Y = \frac{1}{2} \left( \frac{2x+3}{x+2} \right)$

2. Form the cartesian equation of the locus.

Since $X = x$ and $Y = \frac{2x+3}{2(x+2)}$, we can relate the coordinates directly to form the cartesian equation:
$y = \frac{2x+3}{2x+4}$