Past Exam Questions

The equation of a curve is \(\sqrt{x} + \sqrt{y} = \sqrt{a}\), where \(a\) is a positive constant.

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

(ii) The straight line with equation \(y = x\) intersects the curve at the point \(P\). Find the equation of the tangent to the curve at \(P\).

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

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

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

The equation of a curve is \(x^3 + y^3 + 2xy + 8 = 0\).

(a) Express \(\frac{dy}{dx}\) in terms of \(x\) and \(y\).

The tangent to the curve at the point where \(x = 0\) and the tangent at the point where \(y = 0\) intersect at the acute angle \(\alpha\).

(b) Find the exact value of \(\tan \alpha\).

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

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

(b) Find the coordinates of the point on the curve where the tangent is parallel to the \(x\)-axis.

The equation of a curve is \(ye^{2x} - y^2 e^x = 2\).

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

(b) Find the exact coordinates of the point on the curve where the tangent is parallel to the y-axis.

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

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

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

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

The parametric equations of a curve are

\(x = (\ln t)^2\), \(y = e^{2-t^2}\),

for \(t > 0\).

Find the gradient of the curve at the point where \(t = e\), simplifying your answer.

The diagram shows the curve with parametric equations

\(x = \tan \theta, \quad y = \cos^2 \theta\),

for \(-\frac{1}{2}\pi < \theta < \frac{1}{2}\pi\).

(a) Show that the gradient of the curve at the point with parameter \(\theta\) is \(-2 \sin \theta \cos^3 \theta\).

The gradient of the curve has its maximum value at the point \(P\).

(b) Find the exact value of the \(x\)-coordinate of \(P\).

The parametric equations of a curve are

\(x = 3 - \\cos 2\theta\), \(y = 2\theta + \\sin 2\theta\),

for \(0 < \theta < \frac{1}{2}\pi\).

Show that \(\frac{dy}{dx} = \cot \theta\).

The parametric equations of a curve are

\(x = 2t + \\sin 2t, \quad y = \\ln(1 - \\cos 2t)\).

Show that \(\frac{dy}{dx} = \csc 2t\).

The parametric equations of a curve are

\(x = 2 \sin \theta + \sin 2\theta, \quad y = 2 \cos \theta + \cos 2\theta,\)

where \(0 < \theta < \pi\).

1. Obtain an expression for \(\frac{dy}{dx}\) in terms of \(\theta\).
2. Hence find the exact coordinates of the point on the curve at which the tangent is parallel to the \(y\)-axis.

The parametric equations of a curve are

\(x = 2t + \sin 2t, \quad y = 1 - 2 \cos 2t,\)

for \(-\frac{1}{2}\pi < t < \frac{1}{2}\pi\).

(i) Show that \(\frac{dy}{dx} = 2 \tan t.\)

(ii) Hence find the \(x\)-coordinate of the point on the curve at which the gradient of the normal is 2. Give your answer correct to 3 significant figures.

The parametric equations of a curve are

\(x = t^2 + 1, \quad y = 4t + \ln(2t - 1)\).

(i) Express \(\frac{dy}{dx}\) in terms of \(t\).

(ii) Find the equation of the normal to the curve at the point where \(t = 1\). Give your answer in the form \(ax + by + c = 0\).

The parametric equations of a curve are

\(x = \\ln \, \cos \theta\), \(y = 3\theta - \tan \theta\),

where \(0 \leq \theta < \frac{1}{2}\pi\).

(i) Express \(\frac{dy}{dx}\) in terms of \(\tan \theta\).

(ii) Find the exact \(y\)-coordinate of the point on the curve at which the gradient of the normal is equal to 1.

The parametric equations of a curve are

\(x = t + \\cos t\), \(y = \\ln(1 + \\sin t)\),

where \(-\frac{1}{2}\pi < t < \frac{1}{2}\pi\).

(i) Show that \(\frac{dy}{dx} = \sec t\).

(ii) Hence find the \(x\)-coordinates of the points on the curve at which the gradient is equal to 3. Give your answers correct to 3 significant figures.

The parametric equations of a curve are

\(x = a \cos^4 t, \quad y = a \sin^4 t,\)

where \(a\) is a positive constant.

1. Express \(\frac{dy}{dx}\) in terms of \(t\).
2. Show that the equation of the tangent to the curve at the point with parameter \(t\) is \(x \sin^2 t + y \cos^2 t = a \sin^2 t \cos^2 t\).
3. Hence show that if the tangent meets the x-axis at \(P\) and the y-axis at \(Q\), then \(OP + OQ = a\), where \(O\) is the origin.

A curve is defined for \(0 < \theta < \frac{1}{2}\pi\) by the parametric equations

\(x = \tan \theta, \quad y = 2 \cos^2 \theta \sin \theta\).

Show that \(\frac{dy}{dx} = 6 \cos^5 \theta - 4 \cos^3 \theta\).

The parametric equations of a curve are

\(x = \sqrt{t} + 3, \quad y = \ln t\),

for \(t > 0\).

(a) Obtain a simplified expression for \(\frac{dy}{dx}\) in terms of \(t\).

(b) Hence find the exact coordinates of the point on the curve at which the gradient of the normal is \(-2\).

The parametric equations of a curve are

\(x = \frac{1}{\cos^3 t}\), \(y = \tan^3 t\),

where \(0 \leq t < \frac{1}{2} \pi\).

(i) Show that \(\frac{dy}{dx} = \sin t\).

(ii) Hence show that the equation of the tangent to the curve at the point with parameter \(t\) is \(y = x \sin t - \tan t\).