Find equations of all parabolas passing through points (0,3), (2,0) and tangential to $x+y=0$ I need to find equations of all parabolas, that pass through points $(0,3)$ and $(2,0)$ on affine plane $\mathbb{R}^{2}$ and tangent to line $x+y=0$ at origin.
I am trying to solve this problem using conformal geometry. Any hints?

I don't know how to structure an argument using conformal geometry, so here's an alternative approach.
Using a similar strategy to this answer, we note that the equation for the conic through five points $P=(P_x, P_y)$, $Q=(Q_x,Q_y)$, $R=(R_x,R_y)$, $S=(S_x,S_y)$, $T=(T_x,T_y)$ is given by
$$\left|\begin{array}{c,c,c,c,c,c}
x^2 & y^2 & x y & x & y & 1 \\
P_x^2 & P_y^2 & P_x P_y & P_x & P_y & 1 \\ 
Q_x^2 & Q_y^2 & Q_x Q_y & Q_x & Q_y & 1 \\ 
R_x^2 & R_y^2 & R_x R_y & R_x & R_y & 1 \\ 
S_x^2 & S_y^2 & S_x S_y & S_x & S_y & 1 \\ 
T_x^2 & T_y^2 & T_x T_y & T_x & T_y & 1 \\ 
\end{array}\right| = 0 \tag{$\star$}$$
Generalizing the problem at hand a bit, let's take $P:=(p^2,0)$ and $Q:=(0,q^2)$ (the squares are convenient later). Let $R:=(0,0)$. Point $S$ will be a vanishingly-small displacement of $R$ along the tangent line $x+y=0$; so $S := (s,-s)$. Finally, $T$ is some other point; say, $T = (t\cos\theta,t\sin\theta)$. Then $(\star)$ becomes, after dividing through by $s$ and then also taking $s\to 0$:
$$\begin{align}
0 &= x^2 q^2 t \sin\theta\cos\theta + y^2 p^2 t \sin\theta\cos\theta \\
&-x y\left( p^2 t \sin^2\theta + q^2 t \cos^2\theta - p^2 q^2 (\cos\theta+\sin\theta) \right) \\
&- x\,p^2 q^2 t \sin\theta\cos\theta \\
&- y\,p^2 q^2 t \sin\theta\cos\theta  
\end{align} \tag{1}$$
Since our conic is specifically a parabola, we must have
$$(\text{coefficient of}\;xy)^2 = 4(\text{coefficient of}\;x^2)(\text{coefficient of}\;y^2) \tag{2}$$
This condition yields a quadratic in $t$, which we solve and simplify to give
$$t = \frac{p^2 q^2 (\cos\theta + \sin\theta)}{\left(p\sin\theta\pm q\cos\theta\,\right)^2}\tag{3}$$
Substituting back into $(1)$, we discard a common factor of $p^2q^2\sin\theta\cos\theta(\sin\theta+\cos\theta)/(p\sin\theta\pm q\cos\theta)$, leaving an equation that's independent of $T$ except for a sign ambiguity:
$$q^2 x^2 + p^2 y^2 \pm 2p q\,x y - p^2 q^2 x - p^2 q^2 y = 0 \tag{4}$$
which we can write as

$$\left(\frac{x}{p}\pm\frac{y}{q}\right)^2 = x+y
\tag{$\star\star$}$$

Substituting $p\to\sqrt{2}$ and $q\to\sqrt{3}$ is left as an exercise to the reader. $\square$

