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

1. Express the rational function in the form $y = Px + Q + \frac{R}{x + 2}$ using algebraic division or identity.

$\begin{aligned} ax^2 + bx + 3 &\equiv (x + 2)(Px + Q) + R \\ &\equiv Px^2 + (2P + Q)x + 2Q + R \end{aligned}$
Comparing the coefficients:
$P = a$
$2P + Q = b \implies Q = b - 2a$

Thus, $y = ax + (b - 2a) + \frac{R}{x + 2}$.

2. Identify the equation of the oblique asymptote and state the value of $a$.

As $x \to \pm\infty$, the term $\frac{R}{x + 2} \to 0$.
Therefore, the oblique asymptote is $y = ax + (b - 2a)$.
Comparing this with the given asymptote $y = x - 3$, we equate the coefficient of $x$:
$a = 1$

3. Show that $b = -1$ by comparing the constant terms.

Equating the constant terms from the asymptotes:
$b - 2a = -3$
Substitute $a = 1$ into the equation:
$b - 2(1) = -3$
$b - 2 = -3$
$b = -1 \text{ (shown)}$

1. Substitute the values of $a$ and $b$ from part (a) and rearrange the equation to form a quadratic equation in $x$.

$y = \frac{x^2 - x + 3}{x + 2}$
$y(x + 2) = x^2 - x + 3$
$yx + 2y = x^2 - x + 3$
$x^2 - (y + 1)x + (3 - 2y) = 0$

2. Apply the discriminant condition for the quadratic equation to have no real roots, corresponding to values $y$ cannot take.

For $y$ to be a value the function cannot take, there must be no real values of $x$ satisfying the equation. Therefore, the discriminant must be strictly less than zero ($\Delta < 0$):
$b^2 - 4ac < 0$
$(-(y + 1))^2 - 4(1)(3 - 2y) < 0$
$y^2 + 2y + 1 - 12 + 8y < 0$
$y^2 + 10y - 11 < 0$

3. Solve the quadratic inequality to find the set of excluded values.

Factorising the quadratic expression:
$(y + 11)(y - 1) < 0$

Solving the inequality gives:
$-11 < y < 1$

1. Substitute the values of $a$ and $b$ from part (a) to obtain the expression for $y_1$.

From part (a), we have $a = 1$ and $b = -1$. Therefore, the $y$-coordinate of $P$ is:
$y_1 = \frac{x^2 - x + 3}{x + 2}$

2. Apply the midpoint formula to express the $y$-coordinate of the locus in terms of $x$.

Let the coordinates of the midpoint be $(x_m, y_m)$. Since the $x$-coordinates of both $P$ and $Q$ are $x$, $x_m = x$.
The $y$-coordinate of the midpoint is given by $y_m = \frac{y_1 + y_2}{2}$.
$y_m = \frac{1}{2} \left( \frac{x^2 - x + 3}{x + 2} + \frac{3x - x^2}{x + 2} \right)$

3. Simplify the expression to find the final cartesian equation.

$y_m = \frac{1}{2} \left( \frac{x^2 - x + 3 - x^2 + 3x}{x + 2} \right)$
$y_m = \frac{1}{2} \left( \frac{2x + 3}{x + 2} \right)$
$y_m = \frac{2x + 3}{2x + 4}$

Since $x_m = x$, the cartesian equation is $y = \frac{2x + 3}{2x + 4}$.