Finding the curve whose asymptotes are given? 
Find the equation of the cubic curve whose asymptotes are $x+a=0$,
  $y-a=0$, $x+y+a=0$ and which touches the axis of X at origin and
  passes through the point (-2a,-2a)?

My solution: The general equation of the curve must be $(x+a)(y-a)(x+y+a)+g(x,y)=0$ 
Its given that curve touches the x-axis at origin, which implies, x-axis $y=0$ is tangent at origin. So, the lowest degree terms in curve equation must have $y$ as a factor (because, we don't know if $y=0$ is going to be the only tangent at origin.If it was then the only lowest degree factor should have bee $y$). In either case, I don't know how to apply this concept and get $g(x,y)$.
I'm stuck here. I don't know how to proceed further. Final answer must be $xy(x+y)+x(y^2-x^2+xy)-6a^2y=0$ 
EDIT: By the way, can any one kindly suggest a good book that contains such problems?
 A: Consider the most general form of a cubic planar curve $$Ax^3+By^3+Cx^2y+Dxy^2+Ex^2+Fxy+Gy^2+Hx+Ky+L=0.$$
Since we know that the curve must pass through the origin we can conclude that $L=0$. Now using the implicit differentiation we obtain $$y'_x=-\frac{3Ax^2+2Cxy+Dy^2+2Ex+Fy+H}{3By^2+Cx^2+2Dxy+Fx+2Gy+K}=\tag{1}$$ $$-\frac{3A+D\left(\frac{y}{x}\right)^2+2C\left(\frac{y}{x}\right)+\frac{1}{x}\left(2E+F\left(\frac{y}{x}\right)\right)+\frac{1}{x^2}H}{3B\left(\frac{y}{x}\right)^2+C+2D\left(\frac{y}{x}\right)+\frac{1}{x}\left(F+2G\left(\frac{y}{x}\right)\right)+\frac{1}{x^2}K}\tag{2}$$
Since we know that the tangent line at the origin is horizontal we have that $H=0$ from $y'(0)=0$. From the asymptotic behavior for $x\longrightarrow\infty$, $y\longrightarrow a$ (horizontal asymptote) we know $y'\longrightarrow 0$. Using these requirements we must conclude that $A=0$ from formula (2). Exactly the same argument will hold for the vertical asymptote $x=-a$. In this case we will need to compute $$x'_y=-\frac{3B+\ldots}{2X\frac{x}{y}+\ldots}\longrightarrow0\mbox{ as }y\longrightarrow\infty.$$ So conclude that $A=B=0$. Lets look at the slant asymptote $x+y=a$. When $x$ approaches infinity in a way that $y/x\longrightarrow-1$ we must have that $y'_x\longrightarrow -1$. Then we have from 1: $$-1=\frac{D+2C}{C+2D}\Longrightarrow C=D,$$ and we narrow down our search to the curves whose forms fit $$Cxy(x+y)+Ex^2+Fxy+Gy^2+Ky=0.$$ Without loosing generality we can set $C=1$ by rescaling. This is possible since $C=0$ the curve will no longer be cubic. Thus we have $$xy(x+y)+Ex^2+Fxy+Gy^2+Ky=0\tag{3}$$
This is a quadratic equation with respect $x$ or $y$. If we try to resolve (3) for $x$ and $y$ we will obtain $$(G+x)y^2+\mbox{lower order terms in }y=0\tag{4}$$ $$(E+y)x^2+\mbox{lower order terms in }x=0\tag{5}$$ If we want to have a horizontal asymptote at $y=a$ the leading coefficient of [5] must vanish for $y=a$ and we have $E=-a$. Similarly, the presence of the vertical asymptote $x=-a$ is only possible when the leading coefficient of [4] is $0$, i.e. $G=a$ and we obtain more narrow description of all such possible curves: $$xy(x+y)-ax^2+Fxy+ay^2+Ky=0\tag{6}$$ We have not used the fact that $(-2a,-2a)$ must belong to the curve. If we substitute these values into (6), another constrain is revealed: $$-8a^2+2Fa-K=0.\tag{7}$$ We know that $y=a-x$ gets closes and closer to the curve (6) if $x$ is sufficiently large. This means that if we substitute $y=a-x$ into (6) and let $x\longrightarrow\infty$ we should obtain $0$ (this follows from the continuity of the left hand-side of (6)): $$0=\lim_{x\to\infty}\left[x(a-x)(x+a-x)-ax^2-a(a-x)^2+Fx(a-x)+K(a-x)\right]=$$ $$=\lim_{x\to\infty}\left[x^2(-a-F)+x(\ldots)+\ldots\right]\Longrightarrow F=-a\Longrightarrow K=-10a^2\mbox{, follows from (7)}.$$ We arrive to the final form of the necessary cubic curve $$xy(x+y)+a(y^2+xy-x^2)-10a^2y=0.$$
W$_\alpha$ confirms our computations here and here.
