tangent value at a point of an hyperbola

Question

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: $x*(2/pi)*a + (y-d)*(2/pi - d) = -f$

Answer 1

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:

1. $P_1(0,1)$ is on the hyperbola
2. $P_2(\alpha,\alpha)$ is on the hyperbola (we are going to abbreviate $\alpha = \frac{2}{\pi}$ for the time being)
3. 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 $P_2$: