The curve \(C\) has equation \(xy=\frac{1}{2}\). The tangents to the curve \(C\) at the distinct points \(P(p,\frac{1}{2p})\) and \(Q(q,\frac{1}{2q})\), where \(p\) and \(q\) are positive, intersect at \(T\) and the normals to \(C\) at these points intersect at \(N\). Show that \(T\) is the point \[\left(\frac{2pq}{p+q},\frac{1}{p+q}\right).\]

Sketch a diagram to help visualise the information given.

What is the gradient of \(xy=\frac{1}{2}\) at a point? How can we then find the equation of a tangent to the curve at a certain point?

Once we have both tangents, how can we find out where they intersect?

If we know the gradient of a line, what is the gradient of a normal to that line?

In the case \(pq=\frac{1}{2}\), find the coordinates of \(N\). Show (in this case) that \(T\) and \(N\) lie on the line \(y=x\) and are such that the product of their distances from the origin is constant.

The blue curve here is \(xy = \dfrac{1}{2}\). You can vary the value of \(p\) to move the points on the curve.

The tangents at \(P\) and \(Q\) are shown in red, while the normals at these points are in green.

The rectangle has side lengths proportional to \(|OT|\) and \(|ON|\).