It is given that the two curves \[y=4-x^2 \qquad \text{and} \qquad mx=k-y^2,\] where \(m>0\), touch exactly once.

- In each of the following four cases, sketch the two curves on a single diagram, noting the coordinates of any intersections with the axes…

Where do the two curves intersect the axes?

- Now set \(m=12\). Show that the \(x\)-coordinate of any point at which the two curves meet satisfies \[x^4-8x^2+12x+16-k=0.\]

We want an equation in \(x\), so let’s try eliminating \(y\) from the two equations.

Let \(a\) be a value of \(x\) at the point where the curves touch. Show that \(a\) satisfies \[a^3-4a+3=0\] and hence find the three possible values of \(a\).

When the curves touch, what else is equal apart from the coordinate of intersection \(x = a\)?

The equation for \(a\) is a cubic, so can we spot any obvious roots to reduce it to a quadratic?

Derive also the equation \[k=-4a^2+9a+16.\] Which of the four sketches in part (i) arise?

To derive the last equation, could we use both the quartic for \(x\) and the cubic in \(a\)?

Once we have the three values of \(a\) where the curves intersect, can we use the last equation to solve for \(k\)?