This question is a bit concerned with the Tate-Shaferevich group, lets start defining $C$ as $C: X^2- \Delta Y^2=4$
which are generally called as Pell-conics, so all in this question $K$ refers to a number field, so its well known that $Ш(C)\cong Cl(k)^2$ ( where $k$ is the quadratic number field ) . ( The above statement is referred from Prof.Franz Lemmermeyer and his paper on Pell-conics which is here ) . I understood that there is no similar interpretation of the cohomological definition of Shafarevich group on elliptic curves side (I am not sure, if there is one such definition quote a reference) .
So I started working out things on it and searching for the ingredients, but I was successful to the extent in stating that we can prove the following :
There is a map $\alpha : Cl(K)\mapstoШ(K)$ such that $\alpha $ is a homomorphism. And I think we can prove the following way :
Let $O_K$ be the ring of integers of $K$ then for each fractional ideal $I$ of $O_K$, the ideal $IO_H$ is principal, where $H$ is the the maximal abelian unramified extension of the field $K$ ( and we know that its degree is equal to class number of $K$ ) . Therefore, there exists $x \in H^*$ such that $IO_H = xO_H$.
For each $\sigma \in \rm{Gal(H/K)}$, we have $(IO_H)^\sigma = IO_H$ so $(x^\sigma) = (x)$ which means that $\sigma(x)/x \in O^{*}_{L}$ . Therefore, $x\in \bar K^*$ suchthat $\sigma(x)/x \in \bar O^*$ so the map $\sigma \mapsto \sigma(x)/x $ is a cocycle in $Z^1(K,\bar O^{*}_{K})$. If $y$ is another generator of $IO_H$, then $y/x \in O^{*}_{L}$ , so $\sigma(x/y)/(x/y)$ is a coboundary, so it is trivial in $H^1(K,\bar O^{*}_{K})$.
Clearly, the map $\alpha$ that takes $I$ to the cocycle $\alpha(I)(\sigma) = \sigma(x)/x$ is a homomorphism. To show that it induces the desired homomorphism, it is enough to check that if $I = (\beta)$ for $\beta \in K^*$, then $\alpha(I)$ is trivial. Then, we may choose $x =\alpha$ so $\alpha(I)(\sigma)= \sigma(\alpha)/\alpha = \alpha/\alpha= 1$, since $\alpha \in K$. Therefore, the cocycle is trivial.
So I am stuck in proving the isomorphism between the above groups which I will do it once I get the inverse of $\alpha$. But assume that isomorphism is proved and we can write $Cl(K)\cong Ш(K)$.
My question is that, if we have the above isomorphism with us then
" Why can't we write that $Cl(K)\cong Ш_E(K)$ for an elliptic curve $E$ defined over a number field $K$ ? "
So It satisfies the quest for finding a cohomological definition in elliptic curve sense .
P.S : I thank all members who suggested me to read basic things first, after reading basic things that the group suggested, I am sure that I improved a lot and I am ready to ask these types of questions and even eligible to hear answers.
Thank you all.