Sketch the form of the curve \(27y^2=x^3\), …

The graph passes through the point \((0,0)\), and at no other point crosses the \(x\)- or \(y\)-axis.

The graph does not exist for negative \(x\).

The curve will be symmetric about the \(x\)-axis, due to the \(y^2\) term.

We can rearrange the equation for positive \(y\) values as \(y=\frac{1}{\sqrt{27}}x^{3/2}\).

Since \(\dfrac{3}{2}>1\), this will be between the graph of \(y=x\) and \(y=x^2\), so it will look something like this…

Sketch of the curve, with properties described above

… and find the equation of the tangent at the point \(P(12,8)\) on the curve.

Differentiating implicitly, we find that \[54y\frac{dy}{dx}=3x^2,\] and so \[\frac{dy}{dx}=\frac{1}{18}\frac{x^2}{y}.\] Evaluating this at \(P(12,8)\), the gradient of the tangent is \[\frac{1}{18}\frac{12^2}{8}=\frac{144}{144}=1.\] So the tangent has the form \[y=x+c.\]

Since we know that the tangent passes through \((12,8)\), we find that \[8=12+c \Longrightarrow c=-4.\] So the equation of the tangent to the curve at \(P\) is \[y=x-4.\]

Verify this tangent cuts the curve again at the point \(Q(3,-1)\).

For the line \(y=x-4\), we have \[-1=3-4=-1\] and so the line passes through \(Q\).

For the curve, we have \[27y^2=27\times (-1)^2=27\ \text{and}\ x^3=3^3=27.\]

So the curve and the line intersect again at \(Q(3,-1)\).

If \(O\) is the origin, find by integration the area enclosed by the arcs \(OP\), \(OQ\) of the curve and the line \(PQ\). [Integration with respect to \(y\) is recommended.]

Here’s the area we want to find:

Same graph as above but with the area enclosed by OP, OQ and PQ shaded

To find the shaded area, we find the area of the trapezium bounded by the lines \(x=0, y=8\), \(y=-1\) and \(PQ\) and then subtract the area under the curve \(x=3y^{2/3}\) between \(y=-1\) and \(y=8\).

The area of the trapezium is \[\frac{12+3}{2}\times 9=\frac{135}{2}.\]

The area under the curve is \[\begin{align*} \int_{-1}^8 3y^{2/3}\, dy &=\left[\frac{9}{5}y^{5/3}\right]^8_{-1}\\ &=\frac{9\times32}{5}+\frac{9}{5}\\ &=\frac{297}{5}. \end{align*}\]

So the area we want is \[\frac{135}{2}-\frac{297}{5}=\frac{81}{10}.\]