First, substitute \(k = 2\) into the equations:

Line: \(y = 2x + 6\)

Curve: \(y = x^2 + 3x + 4\)

Set the equations equal to find intersection points:

\(x^2 + 3x + 4 = 2x + 6\)

Simplify to: \(x^2 + x - 2 = 0\)

Factorize: \((x - 1)(x + 2) = 0\)

Solutions: \(x = 1\) and \(x = -2\)

Find corresponding \(y\)-values:

For \(x = 1\), \(y = 2(1) + 6 = 8\)

For \(x = -2\), \(y = 2(-2) + 6 = 2\)

Points \(A(1, 8)\) and \(B(-2, 2)\)

Distance \(AB\):

\(AB = \sqrt{(1 - (-2))^2 + (8 - 2)^2} = \sqrt{3^2 + 6^2} = \sqrt{45} = 3\sqrt{5}\)

Midpoint of \(AB\):

\(\left(\frac{1 + (-2)}{2}, \frac{8 + 2}{2}\right) = \left(-\frac{1}{2}, 5\right)\)

The line \(\frac{x}{a} + \frac{y}{b} = 1\) intersects the x-axis at \(P(a, 0)\) and the y-axis at \(Q(0, b)\).

The distance \(PQ\) is given by:

\(PQ = \sqrt{a^2 + b^2} = \sqrt{45}\)

Squaring both sides, we get:

\(a^2 + b^2 = 45 \quad (1)\)

The gradient of the line \(PQ\) is:

\(\text{Gradient} = \frac{0 - b}{a - 0} = -\frac{b}{a} = -\frac{1}{2}\)

Solving for \(b\), we have:

\(\frac{b}{a} = \frac{1}{2}\)

\(b = \frac{a}{2} \quad (2)\)

Substitute equation (2) into equation (1):

\(a^2 + \left(\frac{a}{2}\right)^2 = 45\)

\(a^2 + \frac{a^2}{4} = 45\)

\(\frac{4a^2 + a^2}{4} = 45\)

\(5a^2 = 180\)

\(a^2 = 36\)

\(a = 6\)

Substitute \(a = 6\) back into equation (2):

\(b = \frac{6}{2} = 3\)

Thus, \(a = 6\) and \(b = 3\).

To solve the problem, follow these steps:

1. Find the gradient of \(L_1\) using points \(A(2, 5)\) and \(B(10, 9)\):

\(\text{Gradient of } L_1 = \frac{9 - 5}{10 - 2} = \frac{1}{2}\)

2. Since \(L_2\) is parallel to \(L_1\), its equation is:

\(y = \frac{1}{2}x\)

3. The line \(AC\) is perpendicular to \(L_2\), so its gradient is the negative reciprocal:

\(\text{Gradient of } AC = -2\)

4. Using point \(A(2, 5)\), the equation of line \(AC\) is:

\(y - 5 = -2(x - 2)\)

5. Solve the equations \(y = \frac{1}{2}x\) and \(y - 5 = -2(x - 2)\) simultaneously to find \(C\):

Substitute \(y = \frac{1}{2}x\) into \(y - 5 = -2(x - 2)\):

\(\frac{1}{2}x - 5 = -2(x - 2)\)

\(\frac{1}{2}x - 5 = -2x + 4\)

\(\frac{1}{2}x + 2x = 9\)

\(\frac{5}{2}x = 9\)

\(x = \frac{18}{5} = 3.6\)

\(y = \frac{1}{2}(3.6) = 1.8\)

Thus, \(C(3.6, 1.8)\).

6. Calculate the distance \(AC\):

\(AC = \sqrt{(3.6 - 2)^2 + (1.8 - 5)^2}\)

\(AC = \sqrt{1.6^2 + 3.2^2}\)

\(AC = \sqrt{2.56 + 10.24}\)

\(AC = \sqrt{12.8}\)

\(AC \approx 3.58\)

Step 1: Find the coordinates of D.

The midpoint D of AB is given by:

\(D = \left( \frac{1 + 9}{2}, \frac{3 + (-1)}{2} \right) = (5, 1)\)

Step 2: Write an equation for CD² = 20.

The distance squared from C to D is:

\((x - 5)^2 + (y - 1)^2 = 20\)

Step 3: Find an equation for AC = BC and simplify.

The distances squared are:

\((x - 1)^2 + (y - 3)^2 = (x - 9)^2 + (y + 1)^2\)

Expanding both sides:

\(x^2 - 2x + 1 + y^2 - 6y + 9 = x^2 - 18x + 81 + y^2 + 2y + 1\)

Simplifying gives:

\(-2x - 6y + 10 = -18x + 2y + 82\)

\(16x - 8y = 72\)

\(y = 2x - 9\)

Step 4: Find the possible coordinates of C.

Substitute \(y = 2x - 9\) into \((x - 5)^2 + (y - 1)^2 = 20\):

\((x - 5)^2 + (2x - 10)^2 = 20\)

\((x - 5)^2 + (2x - 10)^2 = 20\)

\(5x^2 - 50x + 105 = 0\)

Solving gives \(x = 3\) or \(x = 7\).

For \(x = 3\), \(y = 2(3) - 9 = -3\).

For \(x = 7\), \(y = 2(7) - 9 = 5\).

Thus, the possible coordinates of C are (3, -3) and (7, 5).

To find the coordinates of \(Q\), substitute \(y = x + 4\) into the curve equation:

\(x + 4 = 2x^2 - 4x + 1\)

Simplify to:

\(2x^2 - 5x - 3 = 0\)

Factorize:

\((2x + 1)(x - 3) = 0\)

Solutions are \(x = -\frac{1}{2}\) and \(x = 3\). Since \(P = (3, 7)\), \(Q = \left( -\frac{1}{2}, \frac{3}{2} \right)\).

For part (iii), find the mid-point of \(AP\):

\(\left( \frac{3 + (-\frac{1}{2})}{2}, \frac{7 + \frac{3}{2}}{2} \right) = (2, 3)\)

Gradient of line \(Q\) to mid-point:

\(\frac{3/2 - 3}{-1/2 - 2} = \frac{-3/2}{-5/2} = \frac{1}{5}\)

Equation of line:

\(y - 3 = -\frac{1}{5}(x - 2)\)