# Tamagawa measure of elliptic curve at a place dividing 2

There's quite a bit of set-up for this, but please bear with it! Let $$K$$ be a number field, and let $$E/K$$ be an elliptic curve. Let $$v$$ be a finite place of $$K$$, and suppose that at $$v$$, $$E$$ has minimal model $$y^2+a_1xy+a_3y = x^3+a_2x^2+a_4x+a_6.$$ Let $$\omega$$ be the corresponding minimal differential; that is $$\omega = \frac{dx}{2y+a_1x+a_3}.$$ Let $$\mu_v$$ be the Haar measure on $$K_v$$ such that $$\mu_v(\mathcal{O}_{K_v})=1$$. Finally, define a measure $$\mu_v(\omega,-)$$ on $$E(K_v)$$ by $$\mu_v(\omega,B)=\int_{B}|\omega|_v=\int_{P\in B}|2y(P)+a_1x(P)+a_3|_v^{-1}d\mu_v(x(P)),$$ where $$B$$ is some Borel set of $$E(K_v)$$, and $$|\cdot|_v$$ is the normalized absolute value corresponding to $$v$$. It's easy enough to show that this is a Haar measure on $$E(K_v)$$.

Right, here's the actual question! When $$v(2)=0$$, I have been able to show that $$\mu_v(\omega,E_1(K_v))=\frac{1}{|k_v|},$$ where $$k_v$$ is the residue field of $$K_v$$ and $$E_1$$ is the kernel of reduction. But I am totally stuck trying to prove it when $$v(2)>0$$. I'm certain it should hold in general. The thing that makes it easier when $$v(2)=0$$ is that you can easily find the value of $$|2y+a_1x+a_3|_v$$ based on the value of $$v(x)$$. And also it's easier to use Hensel's lemma. But when $$v(2)>0$$, that pesky factor of $$2$$ appearing next to the $$y$$ ruins everything, and Hensel's lemma becomes much harder to use. Any ideas how to prove this?