A curve has the equation \(x = y \sqrt { 1 - 2 y }\).
Show that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { \sqrt { 1 - 2 y } } { 1 - 3 y } .$$
The point \(A\) on the curve has \(y\)-coordinate - 1 .
Show that the equation of tangent to the curve at \(A\) can be written in the form
$$\sqrt { 3 } x + p y + q = 0$$
where \(p\) and \(q\) are integers to be found.