Past Exam Questions

The equation of a curve is \(2x^3 - y^3 - 3xy^2 = 2a^3\), where \(a\) is a non-zero constant.

1. Show that \(\frac{dy}{dx} = \frac{2x^2 - y^2}{y^2 + 2xy}\).
2. Find the coordinates of the two points on the curve at which the tangent is parallel to the y-axis.

The equation of a curve is \(x^2(x + 3y) - y^3 = 3\).

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

(ii) Hence find the exact coordinates of the two points on the curve at which the gradient of the normal is 1.

The equation of a curve is \(x^3 y - 3xy^3 = 2a^4\), where \(a\) is a non-zero constant.

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

(ii) Hence show that there are only two points on the curve at which the tangent is parallel to the \(x\)-axis and find the coordinates of these points.

The equation of a curve is \(2x^4 + xy^3 + y^4 = 10\).

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

(ii) Hence show that there are two points on the curve at which the tangent is parallel to the x-axis and find the coordinates of these points.

The equation of a curve is \(xy(x - 6y) = 9a^3\), where \(a\) is a non-zero constant. Show that there is only one point on the curve at which the tangent is parallel to the \(x\)-axis, and find the coordinates of this point.

The equation of a curve is \(x^3 - 3x^2y + y^3 = 3\).

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

(ii) Find the coordinates of the points on the curve where the tangent is parallel to the x-axis.

A curve has equation \(\sin y \ln x = x - 2 \sin y\), for \(-\frac{1}{2}\pi \leq y \leq \frac{1}{2}\pi\).

(i) Find \(\frac{dy}{dx}\) in terms of \(x\) and \(y\).

(ii) Hence find the exact \(x\)-coordinate of the point on the curve at which the tangent is parallel to the \(x\)-axis.

The diagram shows the curve \((x^2 + y^2)^2 = 2(x^2 - y^2)\) and one of its maximum points \(M\). Find the coordinates of \(M\).

The equation of a curve is \(3x^2 + 4xy + 3y^2 = 5\).

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

(b) Hence find the exact coordinates of the two points on the curve at which the tangent is parallel to \(y + 2x = 0\).

A curve has equation \(3e^{2x}y + e^xy^3 = 14\). Find the gradient of the curve at the point \((0, 2)\).

The diagram shows the curve with equation

\(x^3 + xy^2 + ay^2 - 3ax^2 = 0\),

where \(a\) is a positive constant. The maximum point on the curve is \(M\). Find the \(x\)-coordinate of \(M\) in terms of \(a\).

For each of the following curves, find the gradient at the point where the curve crosses the y-axis:

(i) \(y = \frac{1 + x^2}{1 + e^{2x}}\);

(ii) \(2x^3 + 5xy + y^3 = 8\).

The equation of a curve is \(\ln(xy) - y^3 = 1\).

(i) Show that \(\frac{dy}{dx} = \frac{y}{x(3y^3 - 1)}\).

(ii) Find the coordinates of the point where the tangent to the curve is parallel to the y-axis, giving each coordinate correct to 3 significant figures.

The equation of a curve is \(3x^2 - 4xy + y^2 = 45\).

(i) Find the gradient of the curve at the point \((2, -3)\).

(ii) Show that there are no points on the curve at which the gradient is 1.

The equation of a curve is

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

1. Show that \(\frac{dy}{dx} = -\frac{y}{x^2}\).
2. Find the equation of the tangent to the curve at the point where \(y = 1\), giving your answer in the form \(ax + by + c = 0\).

The equation of a curve is \(x^3 - x^2y - y^3 = 3\).

(i) Find \(\frac{dy}{dx}\) in terms of \(x\) and \(y\).

(ii) Find the equation of the tangent to the curve at the point \((2, 1)\), giving your answer in the form \(ax + by + c = 0\).

The equation of a curve is \(xy(x+y) = 2a^3\), where \(a\) is a non-zero constant. Show that there is only one point on the curve at which the tangent is parallel to the \(x\)-axis, and find the coordinates of this point.

The equation of a curve is \(x^3 + 2y^3 = 3xy\).

(i) Show that \(\frac{dy}{dx} = \frac{y - x^2}{2y^2 - x}\).

(ii) Find the coordinates of the point, other than the origin, where the curve has a tangent which is parallel to the \(x\)-axis.

Find the gradient of the curve with equation

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

at the point \((2, 1)\).

The equation of a curve is \(x^2y - ay^2 = 4a^3\), where \(a\) is a non-zero constant.

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

(b) Hence find the coordinates of the points where the tangent to the curve is parallel to the y-axis.