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.



Quadratic twist of an elliptic curve given by non-Weierstrass model
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 $E\_d$.


