To summarize what you did so far, we are given a hyperbola with equation
$$
ax^2 + (y-d)^2 + f = 0
$$
and we want to find $a$, $d$, and $f$ such that:
- $P_1(0,1)$ is on the hyperbola
- $P_2(\alpha,\alpha)$ is on the hyperbola (we are going to abbreviate $\alpha = \frac{2}{\pi}$ for the time being)
- The tangent line to the hyperbola at $P_2$ has slope $-1$.
The first condition gives us:
$$
(1-d)^2 + f = 0 \tag{1}
$$
which means $f = -(1-d)^2$. The second condition gives us:
\begin{gather*}
a \alpha^2 + (\alpha - d)^2 + f = 0 \tag{2}
\end{gather*}
Using implicit differentiation:
$$
2ax + 2(y-d)\frac{dy}{dx} = 0
$$
So at any point $(x,y)$ on the hyperbola
$$
\frac{dy}{dx} = - \frac{a x}{y-d}
$$
Therefore, the third condition gives us an equation:
\begin{gather*}
-1 = - \frac{a\alpha}{\alpha -d} \\\iff
a\alpha = \alpha - d \tag{3}
\end{gather*}
If we substitute $(1)$ and $(3)$ into $(2)$, we have an equation for $d$ alone:
$$
(\alpha - d)\alpha + (\alpha - d)^2 - (1-d)^2 = 0
$$
This looks quadratic in $d$, but actually, the two $d^2$ terms cancel and it's linear. The solution is:
$$
d = \frac{2\alpha^2-1}{3\alpha - 2} = \frac{8-\pi^2}{6\pi - 2\pi^2}
\approx 2.1015
$$
Desmos confirms that the hyperbola with this choice of $d$, $a$, and $f$ passes through $P_1$ and $P_2$, and has the correct tangent at $P_2$:
