2025 RVHS P1 Q7

A curve \(C\) has equation \(x^2 + y^2 - 4y = xy\), where \(x \ge 0\).

1. Show that \((2y - x - 4) \frac{\mathrm{d}y}{\mathrm{d}x} = y - 2x\).[2]

The diagram below shows the curve \(C\).

\(M\) is a point on \(C\) with coordinates \((x, y)\) and \(N\) is a fixed point \((3, 0)\). The area of triangle \(OMN\) is denoted by \(A\).

2. Find \(A\) in terms of \(y\).[1]
3. Show that \(\frac{\mathrm{d}A}{\mathrm{d}x} = \frac{3}{2} \left( \frac{y - 2x}{2y - x - 4} \right)\).[2]

4. Hence find the exact value of \(x\) for which \(A\) is a maximum.

   (You do not need to show that \(A\) is a maximum for the value of \(x\) found.)

   [2]