Homomorphism between elliptic curves $C: y^2=x^3+ax^2+bx$ and $\bar{C}: y^2=x^3-2ax^2+(a^2-4b)x$. I am reading Rational Points on Elliptic Curves by Silverman and Tate. In Section 3.4, Page 76, the authors defined two elliptic curves  elliptic curves $C: y^2=x^3+ax^2+bx$ and $\bar{C}: y^2=x^3-2ax^2+(a^2-4b)x$ and a map $\phi: C\rightarrow \bar{C}$ by
$$\phi(x,y)=\left(\frac{y^2}{x^2}, y(\frac{x^2-b}{x^2})\right).$$
On page 80, they try to show that if $P_1, P_2, P_2 \in C({\mathbb{Q}})$ are colinear, then the image of the three points $\phi(P_1), \phi(P_2), \phi(P_3)\in \bar{C}(\mathbb{Q})$ are also colinear.
Suppose the line passing through $P_1,P_2,P_3$ is $y=\lambda x+\nu$. They claim that the line intersecting $\bar{C}$ is
$$y=\bar{\lambda}x+\bar{\nu}, \text{ where } \bar{\lambda}=\frac{\nu\lambda-b}{\nu} \text{ and } \bar{\nu}=\frac{\nu^2-a\nu\lambda+b\lambda^2}{\nu}.$$
I don't know how they come up with these formula. The exact words in the book are "The line intersecting $\bar{C}$ that we take is". It seems pretty obvious to take. But I couldn't derive them using the known fact. 
Any help will be appreciated!
 A: If I am reading your question correctly, then you are asking how to come up with the formula for the line $\bar{C}$.
Well, one down-to-earth method to gain some intuition would be to look at explicit instances of curves $C$ together with three colinear points $P_1, P_2, P_3$ and then apply $\phi$ to the points. Then one can calculate $\bar{C}$ explicitly, and will see that $\lambda$ relates to $\bar{\lambda}$ and $\nu$ to $\bar{\nu}$ exactly as in their formula. Looking at enough instances one can then infer the general formula.
EDIT (more general explanation):
Let us say that the points $P_i$ has coordinates $(x_i, y_i)$. Because they are colinear we have that $\lambda = \frac{y_j-y_i}{x_j-x_i}$ for $i \not = j$. Now, we can apply $\phi$ and obtain formulas for $\phi(P_i)$ in terms of $x_i$ and $y_i$. Now we can compute the slope $\bar{\lambda}$ for the line between the points $\phi(P_i)$ using the same formula with the new coordinates. 
I have not done the computations, but I am confident that it will match with the formulas in the book. Accordingly with $\bar{\nu}$.
