Quadratic twist of an elliptic curve given by non-Weierstrass model Suppose $f(x)$ is a polynomial of degree 4 with integer coefficients and nonzero discriminant. Let $C$ be the hyperelliptic curve of genus 1 defined by $y^2=f(x)$. If we assume that $C$ has a rational point, then $C$ can be given the structure of an elliptic curve $E$. Now let $d$ be a squarefree integer. Thinking of $C$ as just a hyperelliptic curve, it has a quadratic twist $C_d$ defined by $dy^2=f(x)$; this is not necessarily an elliptic curve, as it might have no rational point. Thinking of $C$ as the elliptic curve $E$, it has a quadratic twist $E_d$, which is an elliptic curve. What is the relation between $E_d$ and $C_d$? They are both in some sense the quadratic twist of $C$, but are not the same curve.
 A: Let's assume the characteristic of the ground field $k$ is not $2$.
If $C$ is of the form $y^2=f(x)$ with $f$ a separable quartic, and $E$ is the Jacobian of $C$ (hence $E$ is an elliptic curve), then the Jacobian of the twist $C_d$ is $E_d$. (This is not hard to deduce from the construction of the Jacobian of a genus $1$ curve. See e.g. chapter 20 in Cassels's Lectures on Elliptic Curves for an accessible account of this.) 
In particular this means that, for any $d$, if $C_d$ has a rational point, then $C_d$ is isomorphic to its own Jacobian, and therefore to $E_d$.
A: Classical invariant theory gives you the answer given by Rene. I add this one only to show how very straightforward it is. We assume the characteristic of the field is not 2 or 3. In this case every elliptic curve $E$ can be written as $y^2 = 4x^3 - g_2 x - g_3$ and the quadratic twist $E_d$ can be written as $y^2 = 4x^3 - d^2 g_2 x - d^3 g_3$.
The genus one curve $C$ that you describe has a Jacobian $E$ which is an elliptic curve. Given your presentation of $C$, it must represent an element of the 2-Selmer group of $E$. Computing its Jacobian is completed in this paper http://math.arizona.edu/~wmc/Research/JacobianFinal.pdf in section 3.1.
In particular, to $f(x) = a_4x^4 + \dots + a_0$ we have two fundamental invariants $I$ - a quadratic form and $J$- a cubic form, each in $a_4, \dots, a_0$. A change of variables shows that $C_d : dy^2 = f(x)$ can be rewritten as $y^2 = df(x)$. Therefore the Jacobian of $C_d$ is exactly $E_d$.
