Past Exam Questions

To find the gradient of the curve, we need to differentiate the given equation implicitly with respect to \(x\).

The equation is \(3e^{2x}y + e^xy^3 = 14\).

Differentiate each term:

1. Differentiate \(3e^{2x}y\) using the product rule:

\(\frac{d}{dx}(3e^{2x}y) = 3e^{2x} \cdot \frac{dy}{dx} + 6e^{2x}y\).

2. Differentiate \(e^xy^3\) using the product rule:

\(\frac{d}{dx}(e^xy^3) = e^xy^3 \cdot \frac{dy}{dx} + 3e^xy^2\).

Combine these results to get the derivative of the left-hand side:

\(6e^{2x}y + 3e^{2x} \frac{dy}{dx} + e^xy^3 \frac{dy}{dx} + 3e^xy^2 = 0\).

Substitute \(x = 0\) and \(y = 2\) into the equation:

\(6e^{0} \cdot 2 + 3e^{0} \cdot \frac{dy}{dx} + e^{0} \cdot 2^3 \cdot \frac{dy}{dx} + 3e^{0} \cdot 2^2 = 0\).

Simplify:

\(12 + 3 \frac{dy}{dx} + 8 \frac{dy}{dx} + 12 = 0\).

Combine like terms:

\(24 + 11 \frac{dy}{dx} = 0\).

Solve for \(\frac{dy}{dx}\):

\(11 \frac{dy}{dx} = -24\).

\(\frac{dy}{dx} = -\frac{24}{11}\).

Thus, the gradient of the curve at the point \((0, 2)\) is \(-\frac{4}{3}\).

To find the \(x\)-coordinate of the maximum point \(M\), we need to differentiate the given equation implicitly with respect to \(x\).

The given equation is \(x^3 + xy^2 + ay^2 - 3ax^2 = 0\).

Differentiating implicitly, we have:

\(3x^2 + y^2 + 2xy \frac{dy}{dx} + 2ay \frac{dy}{dx} - 6ax = 0\).

Rearrange to solve for \(\frac{dy}{dx}\):

\(\frac{dy}{dx}(2xy + 2ay) = 6ax - 3x^2 - y^2\).

Set \(\frac{dy}{dx} = 0\) for the maximum point:

\(6ax - 3x^2 - y^2 = 0\).

Substitute \(y^2 = \frac{3ax^2 - x^3}{x + a}\) into the equation:

\(6ax - 3x^2 - \frac{3ax^2 - x^3}{x + a} = 0\).

Simplify and solve for \(x\):

\(6ax(x + a) - 3x^2(x + a) - (3ax^2 - x^3) = 0\).

\(6ax^2 + 6a^2x - 3x^3 - 3ax^2 + x^3 = 0\).

\(3x^3 - 3ax^2 + 6a^2x = 0\).

Factor out \(x\):

\(x(3x^2 - 3ax + 6a^2) = 0\).

Since \(x = 0\) is not a maximum point, solve:

\(3x^2 - 3ax + 6a^2 = 0\).

Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = -3a\), \(c = 6a^2\):

\(x = \frac{3a \pm \sqrt{(-3a)^2 - 4 \cdot 3 \cdot 6a^2}}{6}\).

\(x = \frac{3a \pm \sqrt{9a^2 - 72a^2}}{6}\).

\(x = \frac{3a \pm \sqrt{-63a^2}}{6}\).

\(x = \frac{3a \pm i\sqrt{63}a}{6}\).

Since we need a real solution, consider the simplified form:

\(x = \sqrt{3a}\).

(i) To find the gradient where the curve crosses the y-axis, substitute \(x = 0\) into the equation. The y-intercept is \(y = \frac{1 + 0^2}{1 + e^{0}} = \frac{1}{2}\).

Differentiate \(y = \frac{1 + x^2}{1 + e^{2x}}\) using the quotient rule:

\(\frac{d}{dx} \left( \frac{1 + x^2}{1 + e^{2x}} \right) = \frac{(1 + e^{2x}) \cdot 2x - (1 + x^2) \cdot 2e^{2x}}{(1 + e^{2x})^2}\).

Substitute \(x = 0\) into the derivative:

\(\frac{(1 + e^{0}) \cdot 0 - (1 + 0^2) \cdot 2e^{0}}{(1 + e^{0})^2} = \frac{-2}{4} = -\frac{1}{2}\).

(ii) For \(2x^3 + 5xy + y^3 = 8\), substitute \(x = 0\) to find \(y\):

\(2(0)^3 + 5(0)y + y^3 = 8 \Rightarrow y^3 = 8 \Rightarrow y = 2\).

Differentiate implicitly:

\(\frac{d}{dx}(2x^3) = 6x^2\), \(\frac{d}{dx}(5xy) = 5y + 5x \frac{dy}{dx}\), \(\frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx}\).

Combine to get:

\(6x^2 + 5y + 5x \frac{dy}{dx} + 3y^2 \frac{dy}{dx} = 0\).

Substitute \(x = 0\), \(y = 2\):

\(0 + 5(2) + 0 \cdot \frac{dy}{dx} + 3(2)^2 \frac{dy}{dx} = 0 \Rightarrow 10 + 12 \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{5}{6}\).

(i) Differentiate \(\ln(xy)\) implicitly: \(\frac{1}{xy}(y + x\frac{dy}{dx}) = \frac{1}{x} + \frac{1}{y}\frac{dy}{dx}\).

Differentiate \(y^3\): \(3y^2\frac{dy}{dx}\).

Set the derivatives equal: \(\frac{1}{x} + \frac{1}{y}\frac{dy}{dx} = 3y^2\frac{dy}{dx}\).

Solve for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{y}{x(3y^3 - 1)}\).

(ii) The tangent is parallel to the y-axis when \(\frac{dy}{dx}\) is undefined, i.e., when \(3y^3 - 1 = 0\).

Solve \(3y^3 - 1 = 0\): \(y = 0.693\).

Substitute \(y = 0.693\) into the original equation: \(\ln(x \times 0.693) - (0.693)^3 = 1\).

Solve for \(x\): \(x = 5.47\).

(i) To find the gradient, we need to differentiate the equation implicitly with respect to \(x\):

\(\frac{d}{dx}(3x^2) - \frac{d}{dx}(4xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(45)\).

This gives \(6x - (4y + 4x\frac{dy}{dx}) + 2y\frac{dy}{dx} = 0\).

Rearrange to find \(\frac{dy}{dx}\):

\(6x - 4y = (4x - 2y)\frac{dy}{dx}\).

\(\frac{dy}{dx} = \frac{6x - 4y}{4x - 2y}\).

Substitute \(x = 2\) and \(y = -3\):

\(\frac{dy}{dx} = \frac{6(2) - 4(-3)}{4(2) - 2(-3)} = \frac{12 + 12}{8 + 6} = \frac{24}{14} = \frac{12}{7}\).

(ii) To show there are no points where the gradient is 1, set \(\frac{dy}{dx} = 1\):

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

This implies \(6x - 4y = 4x - 2y\).

Simplifying gives \(2x = 2y\) or \(y = x\).

Substitute \(y = x\) into the original equation:

\(3x^2 - 4x^2 + x^2 = 45\).

This simplifies to \(0 = 45\), a contradiction.

Therefore, there are no points on the curve where the gradient is 1.

(i) Start with the equation \(x \ln y = 2x + 1\).

Differentiate both sides with respect to \(x\):

\(\frac{d}{dx}(x \ln y) = \frac{d}{dx}(2x + 1)\).

Using the product rule on the left side: \(\ln y + x \cdot \frac{1}{y} \cdot \frac{dy}{dx} = 2\).

Rearrange to solve for \(\frac{dy}{dx}\):

\(x \cdot \frac{1}{y} \cdot \frac{dy}{dx} = 2 - \ln y\).

\(\frac{dy}{dx} = \frac{y(2 - \ln y)}{x}\).

Substitute \(\ln y = \frac{2x + 1}{x}\) from the original equation:

\(\frac{dy}{dx} = \frac{y(2 - \frac{2x + 1}{x})}{x} = -\frac{y}{x^2}\).

(ii) Given \(y = 1\), substitute into the original equation:

\(x \ln 1 = 2x + 1 \Rightarrow 0 = 2x + 1 \Rightarrow x = -\frac{1}{2}\).

Find the gradient at this point using \(\frac{dy}{dx} = -\frac{y}{x^2}\):

\(\frac{dy}{dx} = -\frac{1}{(-\frac{1}{2})^2} = -4\).

The equation of the tangent line is \(y - 1 = -4(x + \frac{1}{2})\).

Simplify to the form \(ax + by + c = 0\):

\(y - 1 = -4x - 2 \Rightarrow y + 4x + 1 = 0\).

(i) Differentiate the equation \(x^3 - x^2y - y^3 = 3\) implicitly with respect to \(x\):

\(\frac{d}{dx}(x^3) = 3x^2\)

\(\frac{d}{dx}(-x^2y) = -2xy - x^2 \frac{dy}{dx}\)

\(\frac{d}{dx}(-y^3) = -3y^2 \frac{dy}{dx}\)

Set the derivative equal to zero:

\(3x^2 - (2xy + x^2 \frac{dy}{dx}) - 3y^2 \frac{dy}{dx} = 0\)

Rearrange to solve for \(\frac{dy}{dx}\):

\(3x^2 - 2xy = (x^2 + 3y^2) \frac{dy}{dx}\)

\(\frac{dy}{dx} = \frac{3x^2 - 2xy}{x^2 + 3y^2}\)

(ii) To find the equation of the tangent at \((2, 1)\), first find the gradient:

Substitute \(x = 2\) and \(y = 1\) into \(\frac{dy}{dx} = \frac{3x^2 - 2xy}{x^2 + 3y^2}\):

\(\frac{dy}{dx} = \frac{3(2)^2 - 2(2)(1)}{(2)^2 + 3(1)^2} = \frac{12 - 4}{4 + 3} = \frac{8}{7}\)

The equation of the tangent line is:

\(y - 1 = \frac{8}{7}(x - 2)\)

Rearrange to the form \(ax + by + c = 0\):

\(7(y - 1) = 8(x - 2)\)

\(7y - 7 = 8x - 16\)

\(8x - 7y - 9 = 0\)

To find where the tangent is parallel to the \(x\)-axis, we need \(\frac{dy}{dx} = 0\).

Start by differentiating the given equation \(xy(x+y) = 2a^3\).

Using the product rule, differentiate \(xy(x+y)\):

\(\frac{d}{dx}[xy(x+y)] = x^2 \frac{dy}{dx} + 2xy + y^2 + 2y \frac{dy}{dx}\).

Set \(\frac{dy}{dx} = 0\) for the tangent to be parallel to the \(x\)-axis:

\(y^2 + 2xy = 0\).

Factor the equation:

\(y(y + 2x) = 0\).

Since \(y = 0\) is not possible (as it would make the original equation undefined), we have:

\(y = -2x\).

Substitute \(y = -2x\) into the original equation:

\(x(-2x)(x - 2x) = 2a^3\).

\(-2x^3 = 2a^3\).

Solve for \(x\):

\(x^3 = -a^3\).

\(x = a\).

Substitute \(x = a\) back into \(y = -2x\):

\(y = -2a\).

Thus, the coordinates of the point are \((a, -2a)\).

(i) Differentiate the equation \(x^3 + 2y^3 = 3xy\) implicitly with respect to \(x\):

\(3x^2 + 6y^2 \frac{dy}{dx} = 3y + 3x \frac{dy}{dx}\).

Rearrange to solve for \(\frac{dy}{dx}\):

\(6y^2 \frac{dy}{dx} - 3x \frac{dy}{dx} = 3y - 3x^2\).

\(\frac{dy}{dx} (6y^2 - 3x) = 3y - 3x^2\).

\(\frac{dy}{dx} = \frac{3y - 3x^2}{6y^2 - 3x} = \frac{y - x^2}{2y^2 - x}\).

(ii) For the tangent to be parallel to the \(x\)-axis, \(\frac{dy}{dx} = 0\).

Set \(y - x^2 = 0\) which gives \(y = x^2\).

Substitute \(y = x^2\) into the original equation:

\(x^3 + 2(x^2)^3 = 3x(x^2)\).

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

\(2x^6 = 2x^3\).

\(x^3(x^3 - 1) = 0\).

\(x^3 = 0\) or \(x^3 = 1\).

\(x = 0\) or \(x = 1\).

For \(x = 1\), \(y = 1^2 = 1\).

Thus, the point is \((1, 1)\).

To find the gradient of the curve at the point \((2, 1)\), we need to differentiate the equation implicitly with respect to \(x\).

The given equation is \(2x^2 - 4xy + 3y^2 = 3\).

Differentiate each term:

\(\frac{d}{dx}(2x^2) = 4x\)

\(\frac{d}{dx}(-4xy) = -4y - 4x\frac{dy}{dx}\)

\(\frac{d}{dx}(3y^2) = 6y\frac{dy}{dx}\)

Substitute these into the differentiated equation:

\(4x - 4y - 4x\frac{dy}{dx} + 6y\frac{dy}{dx} = 0\)

Rearrange to solve for \(\frac{dy}{dx}\):

\(-4x\frac{dy}{dx} + 6y\frac{dy}{dx} = 4y - 4x\)

\((6y - 4x)\frac{dy}{dx} = 4y - 4x\)

\(\frac{dy}{dx} = \frac{4y - 4x}{6y - 4x}\)

Substitute \(x = 2\) and \(y = 1\):

\(\frac{dy}{dx} = \frac{4(1) - 4(2)}{6(1) - 4(2)} = \frac{4 - 8}{6 - 8} = \frac{-4}{-2} = 2\)

Thus, the gradient of the curve at the point \((2, 1)\) is 2.

(a) Differentiate the equation \(x^2y - ay^2 = 4a^3\) implicitly with respect to \(x\):

\(\frac{d}{dx}(x^2y) - \frac{d}{dx}(ay^2) = 0\).

Using the product rule, \(\frac{d}{dx}(x^2y) = 2xy + x^2\frac{dy}{dx}\).

For \(\frac{d}{dx}(ay^2) = 2ay\frac{dy}{dx}\).

Substitute these into the equation:

\(2xy + x^2\frac{dy}{dx} - 2ay\frac{dy}{dx} = 0\).

Rearrange to solve for \(\frac{dy}{dx}\):

\(x^2\frac{dy}{dx} - 2ay\frac{dy}{dx} = -2xy\).

\(\frac{dy}{dx}(x^2 - 2ay) = -2xy\).

\(\frac{dy}{dx} = \frac{2xy}{2ay - x^2}\).

(b) For the tangent to be parallel to the y-axis, \(\frac{dy}{dx}\) is undefined, which occurs when the denominator is zero:

\(2ay - x^2 = 0\).

\(x^2 = 2ay\).

Substitute \(x^2 = 2ay\) into the original equation:

\(2ay^2 - ay^2 = 4a^3\).

\(ay^2 = 4a^3\).

\(y^2 = 4a^2\).

\(y = 2a\) or \(y = -2a\).

For \(y = 2a\), \(x^2 = 2a(2a) = 4a^2\), so \(x = 2a\) or \(x = -2a\).

The points are \((2a, 2a)\) and \((-2a, 2a)\).

(i) Differentiate both sides of \(\sqrt{x} + \sqrt{y} = \sqrt{a}\) with respect to \(x\):

\(\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0\)

Rearrange to find \(\frac{dy}{dx}\):

\(\frac{1}{2\sqrt{y}} \frac{dy}{dx} = -\frac{1}{2\sqrt{x}}\)

\(\frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}}\)

(ii) At the intersection point \(P\), \(y = x\), so substitute into the curve equation:

\(\sqrt{x} + \sqrt{x} = \sqrt{a}\)

\(2\sqrt{x} = \sqrt{a}\)

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

\(x = \frac{1}{4}a\)

Thus, \(P\) is \(\left( \frac{1}{4}a, \frac{1}{4}a \right)\).

The slope of the tangent at \(P\) is \(-\frac{\sqrt{y}}{\sqrt{x}} = -1\).

The equation of the tangent line is:

\(y - \frac{1}{4}a = -1(x - \frac{1}{4}a)\)

\(y = -x + \frac{1}{2}a\)

Rearranging gives:

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

(a) Differentiate the equation \(x^3 + 3x^2y - y^3 = 3\) implicitly with respect to \(x\):

\(3x^2 + 6xy \frac{dy}{dx} + 3x^2 \frac{dy}{dx} - 3y^2 \frac{dy}{dx} = 0\).

Rearrange to solve for \(\frac{dy}{dx}\):

\(6xy \frac{dy}{dx} + 3x^2 \frac{dy}{dx} - 3y^2 \frac{dy}{dx} = -3x^2\).

\(\frac{dy}{dx}(6xy + 3x^2 - 3y^2) = -3x^2\).

\(\frac{dy}{dx} = \frac{x^2 + 2xy}{y^2 - x^2}\).

(b) For the tangent to be parallel to the x-axis, \(\frac{dy}{dx} = 0\).

Set the numerator of \(\frac{dy}{dx} = \frac{x^2 + 2xy}{y^2 - x^2}\) to zero:

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

\(x(x + 2y) = 0\).

So, \(x = 0\) or \(x + 2y = 0\).

If \(x = 0\), substitute into the original equation:

\(3x^2y - y^3 = 3\) becomes \(-y^3 = 3\).

\(y = -\sqrt[3]{3}\).

Point: \((0, -\sqrt[3]{3})\).

If \(x + 2y = 0\), then \(y = -\frac{x}{2}\).

Substitute into the original equation:

\(x^3 + 3x^2(-\frac{x}{2}) - (-\frac{x}{2})^3 = 3\).

\(x^3 - \frac{3x^3}{2} - \frac{x^3}{8} = 3\).

\(-\frac{8x^3}{8} - \frac{12x^3}{8} - \frac{x^3}{8} = 3\).

\(-\frac{21x^3}{8} = 3\).

\(x^3 = -\frac{24}{21}\).

\(x = -2\), \(y = 1\).

Point: \((-2, 1)\).

(a) Differentiate the equation \(x^3 + y^3 + 2xy + 8 = 0\) implicitly with respect to \(x\):

\(3x^2 + 3y^2 \frac{dy}{dx} + 2y + 2x \frac{dy}{dx} = 0\).

Rearrange to solve for \(\frac{dy}{dx}\):

\(3y^2 \frac{dy}{dx} + 2x \frac{dy}{dx} = -3x^2 - 2y\).

\(\frac{dy}{dx} (3y^2 + 2x) = -3x^2 - 2y\).

\(\frac{dy}{dx} = \frac{-3x^2 - 2y}{3y^2 + 2x}\).

(b) Find the gradients at the points \((0, -2)\) and \((-2, 0)\):

At \((0, -2)\), substitute into \(\frac{dy}{dx} = \frac{-3(0)^2 - 2(-2)}{3(-2)^2 + 2(0)} = \frac{4}{12} = \frac{1}{3}\).

At \((-2, 0)\), substitute into \(\frac{dy}{dx} = \frac{-3(-2)^2 - 2(0)}{3(0)^2 + 2(-2)} = \frac{-12}{-4} = 3\).

Use the formula for the tangent of the angle between two lines: \(\tan \alpha = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|\), where \(m_1 = \frac{1}{3}\) and \(m_2 = 3\).

\(\tan \alpha = \left| \frac{\frac{1}{3} - 3}{1 + \frac{1}{3} \times 3} \right| = \left| \frac{-\frac{8}{3}}{2} \right| = \frac{4}{3}\).

(a) Differentiate both sides of \(\ln(x+y) = x - 2y\) with respect to \(x\).

The derivative of the left-hand side using the chain rule is \(\frac{1}{x+y} \left(1 + \frac{dy}{dx}\right)\).

The derivative of the right-hand side is \(1 - 2\frac{dy}{dx}\).

Equate the derivatives: \(\frac{1}{x+y} \left(1 + \frac{dy}{dx}\right) = 1 - 2\frac{dy}{dx}\).

Multiply through by \(x+y\): \(1 + \frac{dy}{dx} = (x+y)(1 - 2\frac{dy}{dx})\).

Expand and rearrange: \(1 + \frac{dy}{dx} = x+y - 2(x+y)\frac{dy}{dx}\).

Combine terms: \(1 + \frac{dy}{dx} + 2(x+y)\frac{dy}{dx} = x+y\).

Factor out \(\frac{dy}{dx}\): \(\frac{dy}{dx}(1 + 2(x+y)) = x+y - 1\).

Solve for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{x+y-1}{2(x+y)+1}\).

(b) For the tangent to be parallel to the \(x\)-axis, \(\frac{dy}{dx} = 0\).

Set \(\frac{x+y-1}{2(x+y)+1} = 0\), which implies \(x+y-1 = 0\).

Thus, \(x+y = 1\).

Substitute \(x+y = 1\) into the original equation: \(\ln(1) = x - 2y\).

Since \(\ln(1) = 0\), we have \(x - 2y = 0\).

Solving \(x+y = 1\) and \(x - 2y = 0\) simultaneously:

From \(x - 2y = 0\), \(x = 2y\).

Substitute into \(x+y = 1\): \(2y + y = 1\).

Thus, \(3y = 1\) and \(y = \frac{1}{3}\).

Substitute \(y = \frac{1}{3}\) into \(x = 2y\): \(x = \frac{2}{3}\).

The coordinates are \(x = \frac{2}{3}, y = \frac{1}{3}\).

(a) Differentiate \(ye^{2x}\) with respect to \(x\):

\(\frac{d}{dx}(ye^{2x}) = 2ye^{2x} + e^{2x} \frac{dy}{dx}\).

Differentiate \(y^2 e^x\) with respect to \(x\):

\(\frac{d}{dx}(y^2 e^x) = 2y e^x \frac{dy}{dx} + y^2 e^x\).

Set the derivative of the left-hand side to zero and solve for \(\frac{dy}{dx}\):

\(2ye^{2x} + e^{2x} \frac{dy}{dx} - (2y e^x \frac{dy}{dx} + y^2 e^x) = 0\).

Rearrange to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{2ye^x - y^2}{2y - e^x}\).

(b) For the tangent to be parallel to the y-axis, \(\frac{dy}{dx}\) is undefined, which occurs when the denominator is zero:

\(2y - e^x = 0 \Rightarrow y = \frac{e^x}{2}\).

Substitute \(y = \frac{e^x}{2}\) into the original equation:

\(\left(\frac{e^x}{2}\right)e^{2x} - \left(\frac{e^x}{2}\right)^2 e^x = 2\).

Simplify to find \(x\):

\(\frac{e^{3x}}{2} - \frac{e^{3x}}{4} = 2 \Rightarrow \frac{e^{3x}}{4} = 2 \Rightarrow e^{3x} = 8\).

\(3x = \ln 8 \Rightarrow x = \frac{1}{3} \ln 8 = \ln 2\).

Substitute back to find \(y\):

\(y = \frac{e^{\ln 2}}{2} = 1\).

The coordinates are \((\ln 2, 1)\).

(a) Differentiate the equation \(x^3 + 3xy^2 - y^3 = 5\) implicitly with respect to \(x\):

\(3x^2 + 3(y^2 + 2xy \frac{dy}{dx}) - 3y^2 \frac{dy}{dx} = 0\).

Rearrange to solve for \(\frac{dy}{dx}\):

\(3x^2 + 3y^2 + 6xy \frac{dy}{dx} - 3y^2 \frac{dy}{dx} = 0\).

\(3x^2 + 3y^2 = (3y^2 - 6xy) \frac{dy}{dx}\).

\(\frac{dy}{dx} = \frac{x^2 + y^2}{y^2 - 2xy}\).

(b) For the tangent to be parallel to the y-axis, \(\frac{dy}{dx}\) is undefined, which occurs when the denominator is zero:

\(y^2 - 2xy = 0\).

\(y(y - 2x) = 0\).

So, \(y = 0\) or \(y = 2x\).

For \(y = 0\), substitute into the original equation:

\(x^3 = 5\), so \(x = \sqrt[3]{5}\).

Point: \(\left( \sqrt[3]{5}, 0 \right)\).

For \(y = 2x\), substitute into the original equation:

\(x^3 + 3x(2x)^2 - (2x)^3 = 5\).

\(x^3 + 12x^3 - 8x^3 = 5\).

\(5x^3 = 5\), so \(x = 1\).

Then \(y = 2 \times 1 = 2\).

Point: (1, 2).

To find where the tangent is parallel to the \(x\)-axis, we need \(\frac{dy}{dx} = 0\).

Differentiate the given equation implicitly:

\(\frac{d}{dx}(2x^2y - xy^2) = \frac{d}{dx}(a^3)\).

Using the product rule, we have:

\(4xy + 2x^2\frac{dy}{dx} - (y^2 + 2xy\frac{dy}{dx}) = 0\).

Simplify to get:

\(4xy + 2x^2\frac{dy}{dx} - y^2 - 2xy\frac{dy}{dx} = 0\).

Combine terms:

\((2x^2 - 2xy)\frac{dy}{dx} = y^2 - 4xy\).

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

\(y^2 - 4xy = 0\).

Factor to get:

\(y(y - 4x) = 0\).

So, \(y = 0\) or \(y = 4x\).

Reject \(y = 0\) because it does not satisfy the original equation unless \(a = 0\), which is not allowed as \(a\) is positive.

Substitute \(y = 4x\) into the original equation:

\(2x^2(4x) - x(4x)^2 = a^3\).

Simplify to get:

\(8x^3 - 16x^3 = a^3\).

\(-8x^3 = a^3\).

\(x^3 = -\frac{a^3}{8}\).

\(x = -\frac{a}{2}\).

Substitute \(x = -\frac{a}{2}\) back into \(y = 4x\):

\(y = 4\left(-\frac{a}{2}\right) = -2a\).

Thus, the \(y\)-coordinate of the point is \(y = -2a\).

To find the gradient \(\frac{dy}{dx}\), we first find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).

\(x = (\ln t)^2\) implies \(\frac{dx}{dt} = 2 \ln t \cdot \frac{1}{t} = \frac{2 \ln t}{t}\).

\(y = e^{2-t^2}\) implies \(\frac{dy}{dt} = -2t e^{2-t^2}\).

The gradient \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-2t e^{2-t^2}}{\frac{2 \ln t}{t}} = -t^2 e^{2-t^2} \cdot \frac{t}{2 \ln t} = -\frac{t^3 e^{2-t^2}}{2 \ln t}\).

Substitute \(t = e\):

\(\frac{dy}{dx} = -\frac{e^3 e^{2-e^2}}{2 \ln e} = -\frac{e^3 e^{2-e^2}}{2}\).

Simplifying gives \(-e^{4-e^2}\).

(a) To find the gradient \(\frac{dy}{dx}\), we first find \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\).

\(\frac{dx}{d\theta} = \sec^2 \theta\)

\(\frac{dy}{d\theta} = -2 \sin \theta \cos \theta\)

Using the chain rule, \(\frac{dy}{dx} = \frac{dy}{d\theta} \div \frac{dx}{d\theta} = \frac{-2 \sin \theta \cos \theta}{\sec^2 \theta} = -2 \sin \theta \cos^3 \theta\).

(b) To find the \(x\)-coordinate of \(P\), we need to find where the derivative \(\frac{dy}{dx}\) is maximized.

Using the product rule, \(\frac{d}{d\theta}(-2 \cos^3 \theta \sin \theta) = -2(\cos^4 \theta + 3 \cos^2 \theta \sin^2 \theta)\).

Setting this derivative to zero gives \(3 \tan^2 \theta = 1\), so \(\tan \theta = \pm \frac{1}{\sqrt{3}}\).

Since \(x = \tan \theta\), the \(x\)-coordinate of \(P\) is \(x = -\frac{1}{\sqrt{3}}\).