A curve is described by the equation (x^2)+4xy+(y^2)+27=0. The tangent to the point P, which lies on the curve, is parallel to the x-axis. Given the x-co-ordinate of P is negative, find the co-ordinates of P.

Firstly, implicitly differentiate the function to find dy/dx in terms of y and x: dy=dx = (-x-2y)/(2x+y). Secondly, set this equal to zero to obtain an expression of y in relation to x or vice versa: x=-2y. Thirdly, input this into the original function to find the x or y coordinate candidates for P: -3y^2+27=0. Therefore the candidates are y=+/-3. Substituting this into either the original equation or the tangent equation will produce candidates for the x-co-ordinate. However, it is easier to insert it into x=-2y. This produces x=-6, or x=6. Given the provided condition, Q(-6,3).