= Solution
Move the nonidentity rational <2-torsion> point to $T=(0,0)$ and complete the square. After a rational change of coordinates, the <elliptic curve> has an integral model
$$
E:y^2=x(x^2+ax+b),\qquad b(a^2-4b)\ne0.
$$
The <two-isogeny formula> gives a partner
$$
E':Y^2=X\bigl(X^2-2aX+b'\bigr),\qquad b'=a^2-4b,
$$
and the <isogeny of elliptic curves> and its <dual isogeny> are
$$
\phi(x,y)=\left(x+a+\frac bx,\ y\left(1-\frac b{x^2}\right)\right),\qquad
\widehat\phi(X,Y)=\left(\frac{X-2a+b'/X}{4},\ \frac{Y(1-b'/X^2)}8\right).
$$
Both extend at the exceptional points; their kernels are the identities and the distinguished <2-torsion> points, and $\widehat\phi\phi=[2]$.
The <two-torsion square-class homomorphism> is $\alpha(O)=1$, $\alpha(T)=[b]$, and $\alpha(x,y)=[x]$ otherwise, with values in $\mathbb Q^*/\mathbb Q^{*2}$. Its kernel is $\widehat\phi E'(\mathbb Q)$. Define $\alpha'$ on the partner in the same way. By the <prime-support bound in two-isogeny descent>, representatives are signed squarefree divisors $d$ of $b$, or of $b'$ for the partner. A class is realized precisely when its <quartic covering in a two-isogeny descent> has a rational solution:
$$
N^2=dU^4+aU^2V^2+\frac bd V^4,\qquad \gcd(U,V)=1.
$$
For $UV\ne0$, this comes from $x=d(U/V)^2$ and $y=dUN/V^3$; the cases $U=0$ or $V=0$ account for the distinguished point and the identity. Test this finite list using signs and local congruences, and exhibit rational solutions for the surviving classes. The <local-to-global gap in isogeny descent> matters: local solubility gives an upper bound from the <isogeny Selmer group>, so a rank determination requires matching actual classes or further descent, not merely declaring every locally soluble cover to have a rational point.
Once the two actual images are determined, the <two-isogeny rank formula> is
$$
\boxed{2^r=\frac{|\alpha E(\mathbb Q)|\,|\alpha'E'(\mathbb Q)|}{4}.}
$$
For completeness, put $A=E(\mathbb Q)$ and $B=E'(\mathbb Q)$. The <Mordell-Weil theorem> gives $[A:2A]=2^r|A[2]|$. The <index of a subgroup> formula through $\widehat\phi B$ gives
$$
[A:2A]=|\alpha A|\,|\alpha'B|\big/[\ker\widehat\phi:\ker\widehat\phi\cap\phi A].
$$
Here $\ker\widehat\phi\cap\phi A=\phi(A[2])$, since $\widehat\phi\phi=[2]$, and this image has size $|A[2]|/2$. The denominator is therefore $4/|A[2]|$, proving the displayed <rank of an elliptic curve> formula.
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