已知曲线的方程为 $y = ax^2 + 4x + 3a - 7$，其中 $a$ 为常数。
直线 $y = 2ax - 5$ 在点 $Q$ 处与该曲线相切。

1. Form the intersection equation of the curve and the line

$ax^2 + 4x + 3a - 7 = 2ax - 5$
$ax^2 + (4 - 2a)x + 3a - 2 = 0$

2. Apply the condition for a tangent (discriminant equals zero)

Since the line is a tangent to the curve, the discriminant $b^2 - 4ac = 0$.
$(4 - 2a)^2 - 4(a)(3a - 2) = 0$

3. Expand and simplify the discriminant equation

$16 - 16a + 4a^2 - 12a^2 + 8a = 0$
$-8a^2 - 8a + 16 = 0$
$a^2 + a - 2 = 0$

4. Solve the quadratic equation for $a$

$(a + 2)(a - 1) = 0$
$a = -2 \text{ or } a = 1$

1. Find the possible values of $a$ from the condition that the line is a tangent to the curve (from part i).

Equating the line and the curve:
$ax^2 + 4x + 3a - 7 = 2ax - 5$
$ax^2 + (4 - 2a)x + 3a - 2 = 0$

Since the line is a tangent, the discriminant is zero:
$(4 - 2a)^2 - 4(a)(3a - 2) = 0$
$16 - 16a + 4a^2 - 12a^2 + 8a = 0$
$-8a^2 - 8a + 16 = 0$
$a^2 + a - 2 = 0$
$(a - 1)(a + 2) = 0$
$a = 1$ or $a = -2$

2. Determine the correct value of $a$ using the given condition on the gradient.

The gradient of the line $y = 2ax - 5$ is $2a$.
Given that the gradient is positive, $2a > 0 \implies a > 0$.
Therefore, $a = 1$.

3. Find the coordinates of the point of contact $Q$.

Substitute $a = 1$ into the intersection equation:
$(1)x^2 + (4 - 2(1))x + 3(1) - 2 = 0$
$x^2 + 2x + 1 = 0$
$(x + 1)^2 = 0$
$x = -1$

Substitute $x = -1$ into the line equation to find $y$:
$y = 2(1)(-1) - 5 = -7$

The coordinates of $Q$ are $(-1, -7)$.