Prove that the curves \(y^2 = 16x\) and \(y = x^2 + 3\) touch at the point \((1,4)\).

To show that the two curves touch one another at \((1,4)\), we’ll show that

this point lies on both curves, and that

their gradients are equal at this point.

For the curve \(y = x^2 + 3\), since \(4 = 1^2 + 3\), the point \((1,4)\) lies on the curve.

We also have that \(y' = 2x \implies y'(1) = 2\).

For the curve defined by \(y^2 = 16x\), we have that \(4^2 = 16 \times 1\), and so the point \((1, 4)\) lies on this curve too.

If \(y^2 = 16x\), then \(y = 4\sqrt{x}=4x^{\frac{1}{2}}\), so \[\begin{align*} \frac{dy}{dx} &= 4\times\frac{1}{2}\times x^{-\frac{1}{2}} \\ &=\frac{2}{\sqrt{x}} \end{align*}\]which gives that \(y'(1)=2\).

Thus at \((1,4), y' = 2\) for both curves, and they do indeed kiss at \((1,4)\).

Alternatively, the curves intersect when \((x^2+3)^2=16x\), or \(x^4+6x^2-16x+9=0\).

This factorises to \((x-1)^2(x^2+2x+9)=0\) (we are looking for a double factor of \((x-1)\), since the question tells us to expect a touching at \(x = 1\)).

Thus there is a double root at \(x = 1\), and so the curves touch at \((1,4)\).

We might note the extra information that \(x^2+2x+9=0\) has no roots, since its discriminant is negative, so \((1,4)\) is the only place these curves coincide.

Prove also that the common tangent at this point forms with the axes a triangle of unit area.

The tangent of the two curves at \((1,4)\) is of the form \(y = mx + c\), where \(m\) is the gradient of both curves at this point, and \(c\) is such that \((1,4)\) lies on the line.

We know that \(m = 2\). As \(c\) satisfies \(4 = m \times 1 + c\), we have \(c = 4 - 2 \times 1 = 2\). So the tangent is \(y = 2x + 2.\)

This intersects the axes at \((-1,0)\) and \((0,2)\), and so the area of the required triangle is \[ \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 2 = 1. \]