Constructing a rational map between two elliptic curves I'm grappling with a problem to construct a rational map between two elliptic curves, and I'm not seeing any way through. The problem

Let $a,b\in K$ and put $a' = -2a$ and $b' = a^2 - 4b$. Let $E/K$ and $E'/K$ be the curves defined by the equations $y^2 = x^3 + ax^2 + x$ and $y^2 = x^3 + a'x^2 + b'x$, respectively. Construct a rational map $\phi: E - - \to E'$ of degree 2, and give the associated embedding of the functions fields.

I've tried a number of different maps, picking some (fairly nice) polynomials for the images of $X$ and $Y$, but so far without success. I'm hoping this isn't a completely messy question and that there's something to grab on to, to help in constructing $\phi$, but so far no luck.
I should add, although I know that $E$ and $E'$ are elliptic curves, this is a course on the arithmetic of modular forms, and we have not developed any arithmetic of elliptic curves, so any solutions that requires some advanced material there may not work out too well (though insight would still be appreciated!)
 A: Alright, firstly, there was a typo, making it impossible :( . The first curve was supposed to be
$$ E : y^2 = x^3 + ax^2 + bx.$$
Secondly, this becomes the decomposition of the multiplication by two isogeny. One can find both morphisms in Silverman's Arithmetic of Elliptic Curves. Namely, one has a map
$$ \phi : E \to E'$$
given by
$$ (x,y) \mapsto \left(\frac{y^2}{x^2},\frac{y(b-x^2)}{x^2}\right),$$
and a map
$$ \hat{\phi} :E'\to E$$
given by
$$ (x,y)\mapsto \left(\frac{y^2}{4x^2}, \frac{y(b'-x^2)}{8x^2}\right).$$
He does not show they are isogenies, but this is not too difficult to see. Suppose we have elements $r,s,u,v\in K$ such that $\phi((r,s)) = (u,v)$. Then
\begin{align*}
 u^3 + a'u^2 + b'u &= u(u^2 + a'u + b')\newline
                   &= \frac{s^2}{r^2}\left(\frac{s^4}{r^4} - 2a\frac{s^2}{r^2} + a^2 - 4b\right)\newline
                   &= \frac{s^2}{r^2}\left(\frac{s^4 - 2as^2r^2 + a^2r^4 - 4br^4}{r^4}\right)\newline
                   &= \frac{s^2}{r^6}((s^2 - ar^2)^2 - 4br^4)\newline
                   &= \frac{s^2}{r^6}((r^3 + br)^2 - 4br^4)\ \ \ \ \ \ \text{as}\ (r,s)\in E/K,\newline
                   &= \frac{s^2}{r^6}(r^6 + 2br^4 + b^2r^2 - 4br^4)\newline
                   &= \frac{s^2}{r^4}(r^4 + 2br^2 + b^2 - 4br^2)\newline
                   &= \frac{s^2}{r^4}(r^2 - b)^2\newline
                   &= \left(\frac{s(r^2 - b)}{r^2}\right)^2 = v^2.
\end{align*}
So for every point (nonzero) $(r,s)$ on $E$, we see $\phi$ maps to a point of $E'$.
