Smooth curve defined over K with genus 1 is always isomorphic, over $\overline K$, to an elliptic curve over K Here, the point is the smooth curve defined over K with genus 1 may not have rational point. But to be an elliptic curve defined over K, the base point must be a rational point.
I tried to use translation map to move some point to a rational point. But the result curve may be not defined over K.
 A: Assuming your question is "why is the statement in your title true?", then this is a situation where the language of varieties gets a bit confusing. The key question to ask yourself is "rational over what field?".
The point is that over an algebraically closed field $\overline{K}$, every curve admits (infinitely many) rational points. You can see this directly by plugging in numbers for one of the coordinates, and then using the algebraic closedness of $\overline{K}$ to show that you can fill in the rest of the coordinates and get a $\overline{K}$-rational point.
Another point is that being "defined over $K$" doesn't mean that you're considering the curve as a "curve over $K$".
In terms of rings, the curve $\text{Spec }\mathbb{Q}(i)[x,y]/(x^2 + y^2 + 1)$ is certainly defined over $\mathbb{Q}$ (since its defining equation $x^2 + y^2 + 1$ has coefficients in $\mathbb{Q}$), but it should not be thought of as a "curve over $\mathbb{Q}$", but rather should be thought of as a curve over $\mathbb{Q}(i)$. For example, one pathology of thinking of it as a curve over $\mathbb{Q}$ is this: if you try to apply $\otimes_{\mathbb{Q}}\overline{\mathbb{Q}}$, you'll find that the result will be disconnected. Note that as a curve over $\mathbb{Q}(i)$, it has a ($\mathbb{Q}(i)$-)rational point.
Thus, upon base-changing your given curve to $\overline{K}$, it becomes obvious that as a curve over $\overline{K}$, it has a rational point, and since it's also genus 1, it's an elliptic curve.
A: There is probably a more direct way to prove this, since it is apparently an exercise in Silverman, but one way to see it is: if $C$ has genus $1$, then its Jacobian $E$ is an elliptic curve, and $C$ and $E$ become isomorphic over any field $L$ over which $C$ obtains a rational point.
