1. The gradient of line AB is \(\frac{q - 3}{p - 8}\).

2. The perpendicular bisector has a gradient of -2, so the gradient of AB is \(\frac{1}{2}\).

3. Set \(\frac{q - 3}{p - 8} = \frac{1}{2}\) and solve for q:

\(q - 3 = \frac{1}{2}(p - 8)\)

\(2q - p = -2\)

4. The midpoint of AB is \(\left( \frac{8 + p}{2}, \frac{3 + q}{2} \right)\).

5. Substitute the midpoint into the equation of the perpendicular bisector:

\(\frac{3 + q}{2} = -2 \left( \frac{8 + p}{2} \right) + 4\)

\(3 + q = -2(8 + p) + 8\)

\(3 + q = -16 - 2p + 8\)

\(q = -11 - 2p\)

6. Solve the system of equations:

\(2q - p = -2\)

\(q = -11 - 2p\)

Substitute \(q = -11 - 2p\) into \(2q - p = -2\):

\(2(-11 - 2p) - p = -2\)

\(-22 - 4p - p = -2\)

\(-5p = 20\)

\(p = -4\)

Substitute \(p = -4\) into \(q = -11 - 2p\):

\(q = -11 - 2(-4)\)

\(q = -11 + 8\)

\(q = -3\)

(i) To find the gradient of line \(AB\), use the formula:

\(m = \frac{y_2 - y_1}{x_2 - x_1}\)

Substitute the coordinates of \(A(-3k - 1, k + 3)\) and \(B(k + 3, 3k + 5)\):

\(m = \frac{(3k + 5) - (k + 3)}{(k + 3) - (-3k - 1)} = \frac{2k + 2}{4k + 4}\)

Simplify:

\(m = \frac{2(k + 1)}{4(k + 1)} = \frac{1}{2}\)

The gradient is \(\frac{1}{2}\), independent of \(k\).

(ii) To find the equation of the perpendicular bisector, first find the midpoint of \(AB\):

\(\text{Midpoint} = \left( \frac{-3k - 1 + k + 3}{2}, \frac{k + 3 + 3k + 5}{2} \right) = \left( \frac{-2k + 2}{2}, \frac{4k + 8}{2} \right)\)

Simplify:

\(\left( -k + 1, 2k + 4 \right)\)

The gradient of the perpendicular bisector is the negative reciprocal of \(\frac{1}{2}\):

\(-2\)

Using the point-slope form of a line equation:

\(y - (2k + 4) = -2(x - (-k + 1))\)

Simplify:

\(y - 2k - 4 = -2(x + k - 1)\) \(y - 2k - 4 = -2x - 2k + 2\) \(y + 2x = 6\)

1. Find the gradient of line \(AB\):

\(\text{Gradient of } AB = \frac{5h - h}{4h + 6 - h} = \frac{4h}{3h + 6}\)

2. The gradient of the perpendicular bisector is \(-\frac{3}{2}\), so:

\(\frac{4h}{3h + 6} \times \left(-\frac{3}{2}\right) = -1\)

3. Solve for \(h\):

\(\frac{4h}{3h + 6} = \frac{2}{3}\)

\(12h = 2(3h + 6)\)

\(12h = 6h + 12\)

\(6h = 12\)

\(h = 2\)

4. Find the midpoint of \(AB\):

\(\left(\frac{5h + 6}{2}, \frac{3h}{2}\right) = (8, 6)\)

5. Substitute the midpoint into the equation of the bisector:

\(3(8) + 2(6) = k\)

\(24 + 12 = k\)

\(k = 36\)

(i) To find the intersection points, set the equations equal:

\(\frac{1}{x} + c = cx - 3\)

Multiply through by \(x\) to clear the fraction:

\(1 + cx = cx^2 - 3x\)

Rearrange to form a quadratic equation:

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

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

Use the quadratic formula \(b^2 - 4ac\) to find the discriminant:

\((c + 3)^2 - 4c = c^2 + 6c + 9 - 4c = c^2 + 2c + 9\)

For real solutions, the discriminant must be non-negative:

\(c^2 + 2c + 9 \geq 0\)

Critical values are \(c = -1\) and \(c = -9\). Thus, \(c \leq -9\) or \(c \geq -1\).

(ii) For tangency, the discriminant must be zero. Substitute \(c = -1\) and \(c = -9\) into the quadratic equation:

For \(c = -1\):

\(-x^2 - 2x - 1 = 0\)

Solving gives \(x = -1\).

For \(c = -9\):

\(-9x^2 + 6x - 1 = 0\)

Solving gives \(x = \frac{1}{3}\).

Part (i):

1. Find the midpoint of AB:

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

2. Calculate the gradient of AB:

\(\frac{9-1}{5-1} = 2\)

3. The gradient of the perpendicular bisector is the negative reciprocal:

\(-\frac{1}{2}\)

4. Use the point-slope form to find the equation:

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

5. Simplify to get:

\(2y = 13 - x\)

Part (ii):

1. Substitute the equation of the perpendicular bisector into the curve equation:

\(6\left( \frac{13-x}{2} \right) = 5x^2 - 18x + 19\)

2. Simplify and solve the quadratic equation:

\(5x^2 - 3x - 4 = 0\)

3. Factorize to find x:

\((x-4)(x+1) = 0\)

4. Solutions are x = 4 and x = -1.

5. Find corresponding y values:

For x = 4, \(y = 4.5\); for x = -1, \(y = 7\).

6. Calculate the distance CD:

\(CD^2 = (4 - (-1))^2 + (4.5 - 7)^2 = 5^2 + (-4.5)^2 = 25 + 6.25 = 31.25\)

7. Therefore, \(CD = \sqrt{\frac{125}{4}}\).