Past Exam Questions

(i) To find the angle \(\angle POQ\) in radians, use the formula for arc length: \(s = r\theta\). Given \(s = 9\) cm and \(r = 5\) cm, we have:

\(9 = 5\theta\)

\(\theta = \frac{9}{5} = 1.8 \text{ rad}\)

(ii) To find the length of \(PT\), consider triangle \(POT\). The angle \(\angle POT\) is half of \(\angle POQ\), so \(\angle POT = \frac{1.8}{2} = 0.9 \text{ rad}\). Using the tangent function:

\(PT = 5 \tan(0.9) \approx 6.30 \text{ cm}\)

(iii) To find the area of the shaded region, first calculate the area of sector \(POQ\):

\(\text{Area of sector} = \frac{1}{2} \times 5^2 \times 1.8 = 22.5 \text{ cm}^2\)

Next, calculate the area of triangle \(POT\):

\(\text{Area of } POT = \frac{1}{2} \times 5 \times 6.30 = 15.75 \text{ cm}^2\)

The shaded area is twice the area of triangle \(POT\) minus the area of the sector:

\(\text{Shaded area} = 2 \times 15.75 - 22.5 = 9.00 \text{ cm}^2\)

(i) To find the perimeter of the shaded region, we first use the Pythagorean theorem in triangle OPT to find OT:

\(OT = \sqrt{OP^2 + PT^2} = \sqrt{5^2 + 12^2} = 13 \text{ cm}\)

Next, find QT:

\(QT = OT - OQ = 13 - 5 = 8 \text{ cm}\)

Calculate the angle POQ using the tangent function:

\(\angle POQ = \tan^{-1}\left(\frac{12}{5}\right) = 1.176 \text{ radians}\)

Find the arc length S:

\(S = r\theta = 5 \times 1.176 = 5.88 \text{ cm}\)

Thus, the perimeter of the shaded region is:

\(5.88 + 12 + 8 = 25.9 \text{ cm}\)

(ii) To find the area of the shaded region, calculate the area of sector POQ:

\(\text{Area of sector} = \frac{1}{2}r^2\theta = \frac{1}{2} \times 5^2 \times 1.176 = 14.7 \text{ cm}^2\)

Calculate the area of triangle OPT:

\(\text{Area of triangle} = \frac{1}{2} \times 12 \times 5 = 30 \text{ cm}^2\)

The area of the shaded region is:

\(30 - 14.7 = 15.3 \text{ cm}^2\)

(i) The area of sector AOB is \(\frac{1}{2}r^2\theta\).

In \(\triangle OAX\), \(AX = r\sin\theta\) and \(OX = r\cos\theta\).

The area of \(\triangle OAX\) is \(\frac{1}{2} \times OX \times AX = \frac{1}{2}r^2\sin\theta\cos\theta\).

Thus, the area of the shaded region AXB is:

\(A = \frac{1}{2}r^2\theta - \frac{1}{2}r^2\sin\theta\cos\theta = \frac{1}{2}r^2(\theta - \sin\theta\cos\theta)\)

(ii) Given \(r = 12\) and \(\theta = \frac{1}{6}\pi\):

\(AX = 12\sin\frac{1}{6}\pi = 6\)

\(OX = 12\cos\frac{1}{6}\pi = 6\sqrt{3}\)

\(BX = OB - OX = 12 - 6\sqrt{3}\)

The length of arc AB is \(r\theta = 12 \times \frac{1}{6}\pi = 2\pi\).

Therefore, the perimeter of the shaded region AXB is:

\(P = AX + BX + \text{arc } AB = 6 + (12 - 6\sqrt{3}) + 2\pi = 18 - 6\sqrt{3} + 2\pi\)

To find the length of AX, we use the tangent of the angle \(\frac{1}{6} \pi\) (half of \(\frac{1}{3} \pi\)) in the right triangle formed by O, A, and X.

\(\tan \frac{1}{6} \pi = \frac{AX}{12}\)

\(\tan \frac{1}{6} \pi = \frac{\sqrt{3}}{3}\)

Thus, \(\frac{\sqrt{3}}{3} = \frac{AX}{12}\)

Solving for \(AX\), we get:

\(AX = 12 \times \frac{\sqrt{3}}{3} = 4\sqrt{3}\)

First, calculate the length of OC and AC using trigonometry. Since \(\angle AOB = \frac{1}{3} \pi\), we have:

\(OC = 12 \cos\left(\frac{1}{3} \pi\right) = 12 \times \frac{1}{2} = 6\)

\(AC = 12 \sin\left(\frac{1}{3} \pi\right) = 12 \times \frac{\sqrt{3}}{2} = 6\sqrt{3}\)

The area of the sector AOB is given by:

\(\frac{1}{2} r^2 \theta = \frac{1}{2} \times 12^2 \times \frac{1}{3} \pi = 24\pi\)

The area of rectangle OCDB is:

\(12 \times 6\sqrt{3} = 72\sqrt{3}\)

The area of triangle OAC is:

\(\frac{1}{2} \times 6 \times 6\sqrt{3} = 18\sqrt{3}\)

The area of the shaded region is the area of the rectangle minus the area of the sector and the triangle:

\(72\sqrt{3} - (24\pi + 18\sqrt{3}) = 54\sqrt{3} - 24\pi\)

Thus, the values of a and b are 54 and 24, respectively.

(i) To find angle AOB, use the tangent formula in the right triangle formed by the radius and the tangent: \(\tan(\frac{x}{2}) = \frac{15}{8} = 1.875\). Solving for \(\frac{x}{2}\), we get \(\frac{x}{2} = 1.081\). Therefore, \(x = 2.16\) radians.

(ii) The perimeter of the shaded region is the sum of the lengths of AT, BT, and the arc AB. The arc length is given by \(r\theta = 8 \times 2.16 = 17.3\) cm. Thus, the perimeter is \(15 + 15 + 17.3 = 47.3\) cm.

(iii) The area of sector AOB is \(\frac{1}{2} r^2 \theta = \frac{1}{2} \times 8^2 \times 2.16 = 69.1\) cm². The area of triangle AOT is \(2 \times \frac{1}{2} \times 8 \times 15 = 120\) cm². The shaded area is \(120 - 69.1 = 50.9\) cm².

The gradient of the line \(l\) is found by rearranging \(2x + y = k\) to \(y = -2x + k\). Thus, the gradient is \(-2\).

For the curve \(xy = 12\), differentiate implicitly: \(x \frac{dy}{dx} + y = 0\).

At \(P(2, 6)\), substitute \(x = 2\) and \(y = 6\) into the differentiated equation:

\(2 \frac{dy}{dx} + 6 = 0\)

\(\frac{dy}{dx} = -\frac{6}{2} = -3\)

The gradient of the tangent at \(P\) is \(-3\).

The angle \(\theta\) between the line and the tangent is given by:

\(\tan \theta = \left| \frac{-3 - (-2)}{1 + (-3)(-2)} \right| = \left| \frac{-1}{1 + 6} \right| = \frac{1}{7}\)

\(\theta = \tan^{-1} \left( \frac{1}{7} \right) \approx 8.1^\circ\)

(i) To find BD, use the formula for the perpendicular from the center to the chord: BD = 9\sin(\pi - 2.4). Calculating gives:

\(BD = 9\sin(\pi - 2.4) = 6.08 \text{ cm}\)

(ii) To find the perimeter of the shaded region, calculate the arc length AB and add the lengths of BD, OD, and OA:

Arc AB = 9 \times 2.4 = 21.6 \text{ cm}

Using Pythagoras or trigonometry, OD = 9\cos(\pi - 2.4) = 6.64 \text{ cm}

Perimeter = 21.6 + 6.08 + 9 + 6.64 = 43.3 \text{ cm}

(iii) To find the area of the shaded region, calculate the area of the sector and subtract the area of triangle OBD:

Area of sector = \(\frac{1}{2} \times 9^2 \times 2.4\)

Area of triangle OBD = \(\frac{1}{2} \times 6.08 \times 6.64\)

Total area = \(\frac{1}{2} \times 81 \times 2.4 - \frac{1}{2} \times 6.08 \times 6.64 = 117 \text{ cm}^2\)

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

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

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

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

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

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

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

Set the numerator of \(\frac{dy}{dx} = -\frac{3x^2 + 6x}{2y + 3}\) to zero:

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

Factorize:

\(3x(x + 2) = 0\).

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

Substitute \(x = 0\) into the original equation:

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

Factorize:

\((y - 1)(y + 4) = 0\).

\(y = 1\) or \(y = -4\).

Coordinates: \((0, 1)\) and \((0, -4)\).

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

\(y^2 + 3y = 0\).

Factorize:

\(y(y + 3) = 0\).

\(y = 0\) or \(y = -3\).

Coordinates: \((-2, 0)\) and \((-2, -3)\).

To find the gradient of the curve \(x^3 + 3xy^2 - y^3 = 1\), we need to differentiate implicitly with respect to \(x\).

The derivative of \(x^3\) is \(3x^2\).

For \(3xy^2\), use the product rule: \(3(y^2 + 2xy\frac{dy}{dx}) = 3y^2 + 6xy\frac{dy}{dx}\).

The derivative of \(y^3\) is \(3y^2\frac{dy}{dx}\).

Substituting these into the equation, we get:

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

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

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

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

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

Substitute \(x = 1\) and \(y = 3\):

\(\frac{dy}{dx} = \frac{-3(1)^2 - 3(3)^2}{6(1)(3) - 3(3)^2}\).

\(\frac{dy}{dx} = \frac{-3 - 27}{18 - 27}\).

\(\frac{dy}{dx} = \frac{-30}{-9} = \frac{10}{3}\).

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

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

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

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

Combine and simplify:

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

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

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

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

(ii) For the tangent to be parallel to the y-axis, \(\frac{dx}{dy} = 0\), which implies \(\frac{dy}{dx}\) is undefined. This occurs when the denominator \(y^2 + 2xy = 0\).

Solve \(y^2 + 2xy = 0\):

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

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

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

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

\(x = a\)

Point: \((a, 0)\)

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

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

\(2x^3 + 8x^3 - 12x^3 = 2a^3\)

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

\(x = -a\)

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

Point: \((-a, 2a)\)

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

The derivative of \(y^3\) is \(3y^2 \frac{dy}{dx}\).

The derivative of \(x^2(x + 3y)\) is \(2x(x + 3y) + x^2(1 + 3\frac{dy}{dx})\).

Equating the derivatives:

\(2x(x + 3y) + x^2(1 + 3\frac{dy}{dx}) - 3y^2 \frac{dy}{dx} = 0\)

Simplify and solve for \(\frac{dy}{dx}\):

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

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

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

(ii) The gradient of the normal is 1, so the gradient of the tangent is -1.

Set \(\frac{dy}{dx} = -1\):

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

\(x^2 + 2xy = -(y^2 - x^2)\)

\(x^2 + 2xy = -y^2 + x^2\)

\(2xy = -y^2\)

\(y = -2x\)

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

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

\(x^2(x - 6x) + 8x^3 = 3\)

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

\(3x^3 = 3\)

\(x^3 = 1\)

\(x = 1\)

\(y = -2\)

First point: \((1, -2)\)

For the second point, solve \(x^2 + 2xy = y^2 - x^2\) with \(y = 0\):

\(x^2 = 3\)

\(x = \sqrt[3]{3}\)

Second point: \((\sqrt[3]{3}, 0)\)

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

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

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

Combine and simplify:

\(3x^2 y + x^3 \frac{dy}{dx} - 9xy^2 \frac{dy}{dx} - 3y^3 = 0\)

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

\(x^3 \frac{dy}{dx} - 9xy^2 \frac{dy}{dx} = 3y^3 - 3x^2 y\)

\(\frac{dy}{dx}(x^3 - 9xy^2) = 3y^3 - 3x^2 y\)

\(\frac{dy}{dx} = \frac{3x^2 y - 3y^3}{9xy^2 - x^3}\)

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

Set the numerator of \(\frac{dy}{dx}\) to zero:

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

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

\(y = 0\) or \(x^2 = y^2\)

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

\(x^3(0) - 3x(0)^3 = 2a^4\), which is not possible.

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

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

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

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

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

\(x^4 = -a^4\), which is not possible.

Substitute \(x = -y\):

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

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

\(2x^4 = 2a^4\)

\(x^4 = a^4\)

\(x = a\) or \(x = -a\)

Thus, the points are \((a, -a)\) and \((-a, a)\).

(i) Differentiate the equation \(2x^4 + xy^3 + y^4 = 10\) implicitly with respect to \(x\):

\(\frac{d}{dx}(2x^4) = 8x^3\).

For \(xy^3\), use the product rule: \(\frac{d}{dx}(xy^3) = y^3 + 3xy^2 \frac{dy}{dx}\).

For \(y^4\), use the chain rule: \(\frac{d}{dx}(y^4) = 4y^3 \frac{dy}{dx}\).

Substitute these into the differentiated equation:

\(8x^3 + y^3 + 3xy^2 \frac{dy}{dx} + 4y^3 \frac{dy}{dx} = 0\).

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

\(3xy^2 \frac{dy}{dx} + 4y^3 \frac{dy}{dx} = -8x^3 - y^3\).

\(\frac{dy}{dx} = -\frac{8x^3 + y^3}{3xy^2 + 4y^3}\).

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

\(-8x^3 - y^3 = 0\) or \(y^3 = -8x^3\).

\(y = -2x\).

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

\(2x^4 + x(-2x)^3 + (-2x)^4 = 10\).

\(2x^4 - 8x^4 + 16x^4 = 10\).

\(10x^4 = 10\).

\(x^4 = 1\).

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

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

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

The points are \((-1, 2)\) and \((1, -2)\).

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

Start by differentiating the given equation \(xy(x - 6y) = 9a^3\).

Using the product rule, differentiate the left-hand side:

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

Equate the derivative to zero for the tangent to be horizontal:

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

Simplify and solve for \(\frac{dy}{dx} = 0\):

\(y(x - 6y) = 0\).

This gives \(y = 0\) or \(x - 6y = 0\).

Since \(y = 0\) does not satisfy the original equation, we use \(x - 6y = 0\).

Substitute \(x = 6y\) into the original equation:

\((6y)y(6y - 6y) = 9a^3\).

\(6y^2 = 9a^3\).

\(y = -a\).

Substitute \(y = -a\) back into \(x = 6y\):

\(x = 6(-a) = -3a\).

Thus, the coordinates are \((-3a, -a)\).

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

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

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

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

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

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

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

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

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

Factor to find \(x(x - 2y) = 0\), giving \(x = 0\) or \(x = 2y\).

Substitute \(x = 0\) into the original equation:

\(y^3 = 3\), so \(y = 1.44\).

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

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

\(8y^3 - 12y^3 + y^3 = 3\).

\(-3y^3 = 3\), so \(y = -1\) and \(x = -2\).

The points are \((-2, -1)\) and \((0, 1.44)\).

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

Using the product rule on the left side, we have:

\(\frac{d}{dx}(\sin y \ln x) = \cos y \frac{dy}{dx} \ln x + \sin y \frac{1}{x}\).

The derivative of the right side is:

\(1 - 2\cos y \frac{dy}{dx}\).

Equating the derivatives, we get:

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

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

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

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

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

Set the numerator of \(\frac{dy}{dx}\) to zero:

\(1 - \frac{\sin y}{x} = 0\).

\(x = \sin y\).

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

\(\sin y \ln \sin y = \sin y - 2\sin y\).

\(\ln \sin y = -1\).

\(\sin y = \frac{1}{e}\).

Thus, \(x = \frac{1}{e}\).

To find the coordinates of \(M\), we start by differentiating both sides of the equation \((x^2 + y^2)^2 = 2(x^2 - y^2)\) with respect to \(x\).

The left-hand side is \((x^2 + y^2)^2\). Using the chain rule, the derivative is:

\(2(x^2 + y^2) \cdot (2x + 2y \frac{dy}{dx})\).

The right-hand side is \(2(x^2 - y^2)\). Its derivative is:

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

Setting the derivatives equal gives:

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

Rearranging terms to solve for \(\frac{dy}{dx}\), we set \(\frac{dy}{dx} = 0\) to find horizontal tangents:

\(2(x^2 + y^2) \cdot 2x = 4x\).

Simplifying gives:

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

Substituting \(x^2 + y^2 = 1\) into the original equation:

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

This simplifies to:

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

Using \(x^2 + y^2 = 1\), we solve for \(x^2\) and \(y^2\):

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

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

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

\(1 = 1 + 2y^2 - 2y^2\).

\(x^2 = \frac{3}{4}\), \(y^2 = \frac{1}{4}\).

Thus, \(x = \frac{1}{2} \sqrt{3}\) and \(y = \frac{1}{2}\).

(a) Differentiate the equation \(3x^2 + 4xy + 3y^2 = 5\) implicitly with respect to \(x\):

\(\frac{d}{dx}(3x^2) = 6x\),

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

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

Set the derivative equal to zero:

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

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

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

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

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

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

(b) The slope of the tangent is \(-2\) (parallel to \(y + 2x = 0\)).

Set \(\frac{dy}{dx} = -2\):

\(-\frac{3x + 2y}{2x + 3y} = -2\),

\(3x + 2y = 4x + 6y\),

\(x = -4y\) or \(y = -\frac{x}{4}\).

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

\(3(-4y)^2 + 4(-4y)y + 3y^2 = 5\),

\(48y^2 - 16y^2 + 3y^2 = 5\),

\(35y^2 = 5\),

\(y^2 = \frac{1}{7}\),

\(y = \pm \frac{1}{\sqrt{7}}\).

For \(y = \frac{1}{\sqrt{7}}\), \(x = -4\left(\frac{1}{\sqrt{7}}\right) = -\frac{4}{\sqrt{7}}\).

For \(y = -\frac{1}{\sqrt{7}}\), \(x = 4\left(\frac{1}{\sqrt{7}}\right) = \frac{4}{\sqrt{7}}\).

The coordinates are \(\left( \frac{4}{\sqrt{7}}, \frac{1}{\sqrt{7}} \right)\) and \(\left( -\frac{4}{\sqrt{7}}, -\frac{1}{\sqrt{7}} \right)\).