To find the values of \(c\) and coordinates of \(P\), we set the equations equal since the line is tangent to the curve:

\(cx^2 + 2x - 3 = 6x - c\)

Rearrange to form a quadratic equation:

\(cx^2 + 2x - 3 - 6x + c = 0\)

\(cx^2 - 4x + (c - 3) = 0\)

For the line to be tangent, the discriminant must be zero:

\(b^2 - 4ac = 0\)

Here, \(a = c\), \(b = -4\), \(c = c - 3\).

\((-4)^2 - 4(c)(c - 3) = 0\)

\(16 - 4c^2 + 12c = 0\)

\(4c^2 - 12c - 16 = 0\)

Divide by 4:

\(c^2 - 3c - 4 = 0\)

Factorize:

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

Thus, \(c = 4\) or \(c = -1\).

For \(c = 4\):

Substitute \(c = 4\) into the quadratic equation:

\(4x^2 - 4x + 1 = 0\)

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

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

Substitute \(x = \frac{1}{2}\) into \(y = 6x - c\):

\(y = 6 \times \frac{1}{2} - 4 = -1\)

Coordinates of \(P\) are \(\left( \frac{1}{2}, -1 \right)\).

For \(c = -1\):

Substitute \(c = -1\) into the quadratic equation:

\(x^2 + 4x + 4 = 0\)

\((x + 2)^2 = 0\)

\(x = -2\)

Substitute \(x = -2\) into \(y = 6x - c\):

\(y = 6 \times (-2) + 1 = -11\)

Coordinates of \(P\) are \((-2, -11)\).

To find the points of intersection, substitute the expression for x from the line equation into the curve equation.

From x + 2y = 9, we have x = 9 - 2y.

Substitute into the curve equation xy + 18 = 0:

\((9 - 2y)y + 18 = 0\)

\(9y - 2y^2 + 18 = 0\)

\(-2y^2 + 9y + 18 = 0\)

Rearrange to form a quadratic equation:

\(2y^2 - 9y - 18 = 0\)

Solving this quadratic equation using the quadratic formula:

\(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

where \(a = 2\), \(b = -9\), \(c = -18\).

Calculate the discriminant:

\(b^2 - 4ac = (-9)^2 - 4 \times 2 \times (-18) = 81 + 144 = 225\)

Thus,

\(y = \frac{9 \pm 15}{4}\)

So, \(y = 6\) or \(y = -1.5\).

Substitute back to find \(x\):

For \(y = 6\):

\(x = 9 - 2(6) = -3\)

For \(y = -1.5\):

\(x = 9 - 2(-1.5) = 12\)

Thus, the coordinates of the points are (12, -1.5) and (-3, 6).

To find the values of \(k\) and \(a\), we equate the equations of the curve and the line at the given \(x\)-coordinates.

1. For \(x = 0\):

\(4(0)^2 - k(0) + \frac{1}{2}k^2 = 0 - a\)

\(\frac{1}{2}k^2 = -a\) (Equation 1)

2. For \(x = \frac{3}{4}\):

\(4\left(\frac{3}{4}\right)^2 - k\left(\frac{3}{4}\right) + \frac{1}{2}k^2 = \frac{3}{4} - a\)

\(4\left(\frac{9}{16}\right) - \frac{3}{4}k + \frac{1}{2}k^2 = \frac{3}{4} - a\)

\(\frac{9}{4} - \frac{3}{4}k + \frac{1}{2}k^2 = \frac{3}{4} - a\) (Equation 2)

Substitute \(a = -\frac{1}{2}k^2\) from Equation 1 into Equation 2:

\(\frac{9}{4} - \frac{3}{4}k + \frac{1}{2}k^2 = \frac{3}{4} + \frac{1}{2}k^2\)

\(\frac{9}{4} - \frac{3}{4}k = \frac{3}{4}\)

\(\frac{9}{4} - \frac{3}{4} = \frac{3}{4}k\)

\(\frac{6}{4} = \frac{3}{4}k\)

\(k = 2\)

Substitute \(k = 2\) back into Equation 1:

\(\frac{1}{2}(2)^2 = -a\)

\(2 = -a\)

\(a = -2\)

Thus, \(k = 2\) and \(a = -2\).

To find the values of \(m\), equate the line and curve equations: \(mx - 6 = x^2 - 4x + 3\).

Rearrange to form a quadratic: \(x^2 - (4 + m)x + 9 = 0\).

For the line to be tangent, the discriminant must be zero: \(b^2 - 4ac = 0\).

Here, \(a = 1\), \(b = -(4 + m)\), \(c = 9\).

So, \((4 + m)^2 - 4 \times 1 \times 9 = 0\).

\((4 + m)^2 = 36\).

\(4 + m = 6\) or \(4 + m = -6\).

\(m = 2\) or \(m = -10\).

For \(m = 2\), substitute back to find \(x\):

\(x^2 - 6x + 9 = 0\) gives \((x - 3)^2 = 0\), so \(x = 3\).

Substitute \(x = 3\) into \(y = mx - 6\):

\(y = 2 \times 3 - 6 = 0\).

Point is \((3, 0)\).

For \(m = -10\), substitute back to find \(x\):

\(x^2 + 6x + 9 = 0\) gives \((x + 3)^2 = 0\), so \(x = -3\).

Substitute \(x = -3\) into \(y = mx - 6\):

\(y = -10 \times (-3) - 6 = 24\).

Point is \((-3, 24)\).

To find the \(x\)-coordinates of the intersection points, set the equations equal: \(7\sqrt{x} = 6x + 2\).

Rearrange to form a quadratic equation: \(6x + 2 = 7\sqrt{x} \Rightarrow 6(\sqrt{x})^2 - 7\sqrt{x} + 2 = 0\).

Let \(t = \sqrt{x}\), then the equation becomes \(6t^2 - 7t + 2 = 0\).

Factor the quadratic: \((3t - 2)(2t - 1) = 0\).

Solving for \(t\), we get \(t = \frac{2}{3}\) or \(t = \frac{1}{2}\).

Since \(t = \sqrt{x}\), we have \(\sqrt{x} = \frac{2}{3}\) or \(\sqrt{x} = \frac{1}{2}\).

Thus, \(x = \left(\frac{2}{3}\right)^2 = \frac{4}{9}\) or \(x = \left(\frac{1}{2}\right)^2 = \frac{1}{4}\).

Given the equations:

1. \(y^2 + 2x = 13\)

2. \(2y + x = 8\)

From equation (2), express \(x\) in terms of \(y\):

\(x = 8 - 2y\)

Substitute \(x = 8 - 2y\) into equation (1):

\(y^2 + 2(8 - 2y) = 13\)

Simplify:

\(y^2 + 16 - 4y = 13\)

\(y^2 - 4y + 3 = 0\)

Factor the quadratic equation:

\((y - 3)(y - 1) = 0\)

Thus, \(y = 3\) or \(y = 1\).

For \(y = 3\):

\(x = 8 - 2(3) = 2\)

Point: \((2, 3)\)

For \(y = 1\):

\(x = 8 - 2(1) = 6\)

Point: \((6, 1)\)

Therefore, the points of intersection are \((2, 3)\) and \((6, 1)\).

Given the equations:

Line: \(2y = x + 5\)

Curve: \(y = x^2 - 4x + 7\)

Substitute \(y\) from the line equation into the curve equation:

\(y = \frac{x + 5}{2}\)

Substitute into the curve equation:

\(\frac{x + 5}{2} = x^2 - 4x + 7\)

Multiply through by 2 to eliminate the fraction:

\(x + 5 = 2x^2 - 8x + 14\)

Rearrange to form a quadratic equation:

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

Factor or use the quadratic formula to solve for \(x\):

\(x = 3\) or \(x = \frac{1}{2}\)

(i) To find the points of intersection, substitute \(y = 3 - x\) from the line equation into the curve equation:

\(2x^2 - 8x + 9 = 3 - x\)

Rearrange to form a quadratic equation:

\(2x^2 - 7x + 6 = 0\)

Factorize:

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

Thus, \(x = 2\) or \(x = \frac{3}{2}\).

(ii) To find the stationary point of C, differentiate \(y = 2x^2 - 8x + 9\):

\(\frac{dy}{dx} = 4x - 8\)

Set \(\frac{dy}{dx} = 0\):

\(4x - 8 = 0\)

\(x = 2\)

Therefore, \(x = 2\) is a stationary point of C.

Given the equations:

1. Curve: \(xy = 12\)

2. Line: \(2x + y = 11\)

Substitute \(y = 11 - 2x\) from the line equation into the curve equation:

\(x(11 - 2x) = 12\)

\(11x - 2x^2 = 12\)

Rearrange to form a quadratic equation:

\(2x^2 - 11x + 12 = 0\)

Factor the quadratic equation:

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

Thus, \(x = \frac{3}{2}\) or \(x = 4\).

For \(x = \frac{3}{2}\), substitute back to find \(y\):

\(y = 11 - 2 \times \frac{3}{2} = 8\)

For \(x = 4\), substitute back to find \(y\):

\(y = 11 - 2 \times 4 = 3\)

Therefore, the points of intersection are \((1.5, 8)\) and \((4, 3)\).

To find the points of intersection, substitute \(y = 11 - 2x\) from the line equation into the curve equation \(xy = 12\).

Substitute: \(x(11 - 2x) = 12\).

Expand and rearrange: \(11x - 2x^2 = 12\).

Rearrange to form a quadratic equation: \(2x^2 - 11x + 12 = 0\).

Factor the quadratic: \((2x - 3)(x - 4) = 0\).

Solve for \(x\): \(x = \frac{3}{2}\) or \(x = 4\).

Substitute back to find \(y\):

For \(x = \frac{3}{2}\), \(y = 11 - 2 \times \frac{3}{2} = 8\).

For \(x = 4\), \(y = 11 - 2 \times 4 = 3\).

Thus, the points of intersection are \(\left( \frac{1}{2}, 8 \right)\) and \((4, 3)\).

(i) The coordinates of \(P\) are \((t, 5t)\) since it lies on the line \(y = 5x\). The coordinates of \(Q\) are \((t, t(9 - t^2))\) since it lies on the curve \(y = x(9 - x^2)\). The length of \(PQ\) is the difference in the \(y\)-coordinates of \(Q\) and \(P\):

\(PQ = t(9 - t^2) - 5t = 9t - t^3 - 5t = 4t - t^3\)

(ii) To find the maximum value of \(PQ\), differentiate \(PQ = 4t - t^3\) with respect to \(t\):

\(\frac{d(PQ)}{dt} = 4 - 3t^2\)

Set the derivative to zero to find critical points:

\(4 - 3t^2 = 0\)

\(3t^2 = 4\)

\(t^2 = \frac{4}{3}\)

\(t = \frac{2}{\sqrt{3}}\)

Substitute \(t = \frac{2}{\sqrt{3}}\) back into \(PQ = 4t - t^3\) to find the maximum length:

\(PQ = 4\left(\frac{2}{\sqrt{3}}\right) - \left(\frac{2}{\sqrt{3}}\right)^3\)

\(PQ = \frac{8}{\sqrt{3}} - \frac{8}{3\sqrt{3}}\)

\(PQ = \frac{24}{3\sqrt{3}} - \frac{8}{3\sqrt{3}}\)

\(PQ = \frac{16}{3\sqrt{3}}\)

Alternatively, \(PQ = \frac{16\sqrt{3}}{9}\).

(i) To find the stationary point B, we set the derivative of the curve to zero: \(\frac{dy}{dx} = 2x - 4 = 0\). Solving gives \(x = 2\). Substituting \(x = 2\) into the curve equation gives \(y = 3\). Thus, B is (2, 3).

The midpoint of AB is \(\left( \frac{4+2}{2}, \frac{7+3}{2} \right) = (3, 5)\).

Since L passes through this midpoint, substitute (3, 5) into \(y = mx - 2\):

\(5 = 3m - 2\)

Solving for \(m\), we get \(m = \frac{7}{3}\).

(ii) The line L does not meet the curve if the quadratic equation formed by substituting \(y = mx - 2\) into the curve equation has no real solutions. This occurs when the discriminant is less than zero.

Substitute \(y = mx - 2\) into \(y = x^2 - 4x + 7\):

\(x^2 - 4x + 7 = mx - 2\)

Rearrange to form a quadratic: \(x^2 - (4+m)x + 9 = 0\)

The discriminant \(b^2 - 4ac\) is \((m+4)^2 - 36\).

Set \((m+4)^2 - 36 < 0\) and solve:

\((m+4)^2 < 36\)

\(-6 < m+4 < 6\)

\(-10 < m < 2\)

(i) To find the points of intersection, set the equations equal: \(x^2 - x + 3 = 3x + a\). Rearrange to get \(x^2 - 4x + (3 - a) = 0\).

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

\((-1)^2 - 4(-1) + (3 - a) = 0\)

\(1 + 4 + 3 - a = 0\)

\(8 - a = 0 \Rightarrow a = 8\)

Substitute \(a = 8\) back into the equation:

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

Factorize: \((x - 5)(x + 1) = 0\)

The other \(x\)-coordinate is \(x = 5\).

(iii) For the line to be tangent, the discriminant must be zero:

\(b^2 - 4ac = 0\)

\((-4)^2 - 4(1)(3 - a) = 0\)

\(16 - 4(3 - a) = 0\)

\(16 - 12 + 4a = 0\)

\(4a = -4 \Rightarrow a = -1\)

Substitute \(a = -1\) into \(x^2 - 4x + 4 = 0\):

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

Substitute \(x = 2\) into \(y = x^2 - x + 3\):

\(y = 2^2 - 2 + 3 = 5\)

The coordinates of \(P\) are \((2, 5)\).

(i) Equate the line and curve equations: \(-2x + k = \frac{2}{x - 3}\).

Multiply through by \(x - 3\) to clear the fraction: \(-2x(x - 3) + k(x - 3) = 2\).

Simplify to get \(2x^2 - (6 + k)x + (2 + 3k) = 0\).

(ii) For the line to be tangent to the curve, the quadratic equation must have a repeated root, so the discriminant \(b^2 - 4ac = 0\).

Here, \(a = 2\), \(b = -(6 + k)\), \(c = 2 + 3k\).

Calculate the discriminant: \((6 + k)^2 - 4(2)(2 + 3k) = 0\).

Simplify: \(k^2 - 12k + 20 = 0\).

Factorize: \((k - 10)(k - 2) = 0\).

Thus, \(k = 2\) or \(k = 10\).

(iii) For \(k = 2\), substitute into the quadratic: \(2(x - 2)^2 = 0\).

\(x = 2\), \(y = -2\), so \(A = (2, -2)\).

For \(k = 10\), substitute into the quadratic: \(2(x - 4)^2 = 0\).

\(x = 4\), \(y = 2\), so \(B = (4, 2)\).

Find the equation of line \(AB\):

Slope \(m = \frac{2 - (-2)}{4 - 2} = 2\).

Using point-slope form: \(y + 2 = 2(x - 2)\) or \(y = 2x - 6\).

(i) To find the \(x\)-coordinates of \(A\) and \(B\), set the equations equal: \(2x^5 + 3x^3 = 2x\).

Rearrange to: \(2x^5 + 3x^3 - 2x = 0\).

Factor out \(x\): \(x(2x^4 + 3x^2 - 2) = 0\).

Thus, \(2x^4 + 3x^2 - 2 = 0\) is satisfied by the \(x\)-coordinates of \(A\) and \(B\).

(ii) Solve \(2x^4 + 3x^2 - 2 = 0\) by substituting \(y = x^2\), giving \(2y^2 + 3y - 2 = 0\).

Factorize: \((y + 2)(2y - 1) = 0\).

Thus, \(y = -2\) or \(y = \frac{1}{2}\).

Since \(y = x^2\), \(x^2 = \frac{1}{2}\) (ignore \(y = -2\) as \(x^2\) cannot be negative).

So, \(x = \pm \frac{1}{\sqrt{2}}\).

For \(x = \frac{1}{\sqrt{2}}\), \(y = 2x = \frac{2}{\sqrt{2}}\).

For \(x = -\frac{1}{\sqrt{2}}\), \(y = 2x = -\frac{2}{\sqrt{2}}\).

Thus, the coordinates are \(\left( \frac{1}{\sqrt{2}}, \frac{2}{\sqrt{2}} \right)\) and \(\left( -\frac{1}{\sqrt{2}}, -\frac{2}{\sqrt{2}} \right)\).

(i) To show the curve lies above the \(x\)-axis, find the vertex of the parabola. The derivative is \(\frac{dy}{dx} = 2x - 3\). Setting \(\frac{dy}{dx} = 0\) gives \(x = 1.5\). Substituting \(x = 1.5\) into the equation gives \(y = 1.75\). Since the minimum point \(y = 1.75 > 0\), the curve lies above the \(x\)-axis.

(ii) The function is decreasing where \(\frac{dy}{dx} < 0\). Solving \(2x - 3 < 0\) gives \(x < 1.5\).

(iii) Substitute \(y = 6 - 2x\) into \(y = x^2 - 3x + 4\) to get \(x^2 - x - 2 = 0\). Solving gives \(x = -1\) and \(x = 2\). Substituting back gives points \((-1, 8)\) and \((2, 2)\).

(iv) For the line to be tangent, the discriminant of \(x^2 - 3x + 4 - k + 2x = 0\) must be zero. Simplifying gives \(x^2 - x + (4 - k) = 0\). The discriminant is \(b^2 - 4ac = 1^2 - 4(1)(4-k) = 0\). Solving gives \(k = \frac{3}{4}\).