2009 AJC Promo Q13

The curve \(C\) has the equation \(y=\frac{-{{x}^{2}}+ax}{x-2b}\), where \(a\) and \(b\) are non-zero constants. Given that \(y=-x+1\) is an asymptote, find \(a\) in terms of \(b\).[2]

1. Find the range of values of \(b\), such that the curve \(C\) has \(2\) stationary points.[3]
2. Given that \(1 < b < 3\), sketch the curve \(C\). Show clearly (if any), the equations of the asymptotes, and the coordinates of the axial intercepts and stationary points, in terms of \(b\).[4]
3. Show that the intersection point of the asymptotes lies on the line \(y=m(x-2b)+(1-2b)\), where \(m\) is a constant. Hence, find the range of values of \(m\) for which the line \(y=m(x-2b)+(1-2b)\) cuts \(C\) at \(2\) distinct points.[2]