tangent value at a point of an hyperbola I have an hyperbola (centered to the Y axis, displaced in Y) passing through the points $P(X,Y)$, $P_1(0,1)$ $P_2(2/pi,2/pi)$.
This hyperbola has for equation $a*X² + (Y-d)² +f = 0$
So far so good: we can find $a$ and $f$ ($f=(1-d)² $..) depending of $d$ because it pass through $P_1$ and $P_2$ . See graph
https://www.desmos.com/calculator/egiica6qvj
By adding the condition: tangent at$ P_2$ should be -1, we should be able to find $ d$ and not only guess it is around 2.2
Update final: The equation of the tangent line is: $x*(2/pi)*a + (y-d)*(2/pi - d) = -f$
 A: 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 $P_2$:

