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.

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 $E_d$. 


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 2Selmer 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 $E_d$. 


Maple can compute Weierstrass model without a point. Here are some experiments for $f(x)=x^4+a$ which are related to congruent numbers.
$$ E: {v}^{2}+{u}^{3}4\,au = 0$$ $$ x= 2\,{\frac {{\it RootOf} \left( a+{{\it \_Z}}^{2} \right) v}{4\,a{u}^{2}}} $$ $$ y= 1/2\,{\frac {2\,{\it RootOf} \left( a+{{\it \_Z}}^{2} \right) {u}^{2}+8\,{\it RootOf} \left( a+{{\it \_Z}}^{2} \right) a}{4\,a+{u}^{2}}} $$
$$ E_d: {v}^{2}+{u}^{3}4\,{d}^{2}au = 0 $$ $$ x=2\,{\frac {{\it RootOf} \left( {{\it \_Z}}^{2}da \right) v}{4\,{d}^{2}a{u}^{2}}} $$ $$ y= 1/2\,{\frac {8\,{\it RootOf} \left( {{\it \_Z}}^{2}da \right) {d}^{2}a+2\,{\it RootOf} \left( {{\it \_Z}}^{2}da \right) {u}^{2}}{4\,{d}^{3}a+d{u}^{2}}} $$ Observe that you must twist by $1$ and change the sign of $u$. Experiments suggest that if $C$ is of positive rank $E$ has a generator in the number field which maps to rational point. If $a$ is square, the maps are over the rationals. For $f(x)=fx^4+ex^3+c x^2+b x+a$. $$ E: {u}^{3}+ \left( 1/3\,{c}^{2}4\,fa+eb \right) u{\frac {2}{27}}\,{c}^{3}f{b}^{2}{e}^{2}a+8/3\,fca+1/3\,bec+{v}^{2} = 0 $$ $$ E_d: {u}^{3}+ \left( 1/3\,{d}^{2}{c}^{2}4\,{d}^{2}fa+{d}^{2}eb \right) u{\frac {2}{27}}\,{d}^{3}{c}^{3}{d}^{3}{e}^{2}a+1/3\,b{d}^{3}ec{d}^{3}f{b}^{2}+8/3\,{d}^{3}cfa+{v}^{2} = 0 $$


