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?
 A: Since the parabola passes the point $(0,3)$, assume that it takes the form,
$$y=ax^2+bx+3\tag{1}$$
Plug in the point $(2,0)$, we have
$$0=4a+2b+3\tag{2}$$
Substitute $y$ in (1) with the tangential line $y=-x$,
$$ax^2+(1+b)x+3=0$$
Given that the line is tangential to the parabola, the discriminate of the above equation is zero,
$$(1+b)^2 - 12a=0$$
Together with (2), we obtain
$$b=-4\pm \sqrt 6$$
$$a = \frac{5 \mp 2\sqrt 6}{4}$$
There are two solutions for the parabola,
$$y=\frac{5 \pm 2\sqrt 6}{4}x^2 - (4\pm \sqrt 6)x+ 3$$
A: Parabola is a curve that has equal distance from given point $A$ and line $L$. Since parabola pass the origin, so that line can't, so we can suppose $L: ax+by=1$ and $A: (c,d)$. Now write the equation explicitly:
$$\frac{(ax_0+by_0-1)^2}{a^2+b^2}=(x_0-c)^2+(y_0-d)^2$$
We have $3$ points:
$$\left\{\begin{array}{c} \frac{(3b-1)^2}{a^2+b^2}=c^2+(d-3)^2 \\ \frac{(2a-1)^2}{a^2+b^2}=(c-2)^2+d^2 \\ \frac{(-1)^2}{a^2+b^2}=c^2+d^2 \\ \end{array}\right\}$$
It tangent to $x+y=0$ at origin:
$$-\frac{2(a^2+b^2)(x_0-c)-2a(ax_0+by_0-1)}{2(a^2+b^2)(y_0-d)-2b(ax_0+by_0-1)}(x_0=0,y_0=0)=-1 \Longrightarrow$$
$$\Longrightarrow (a^2+b^2)(d-c)=(b-a)$$
So we get $4$ equation of degree $2$ on $4$ variables that you can solve it on:
https://www.wolframalpha.com/widgets/view.jsp?id=ae438682ce61743f90d4693c497621b7
And fine $2$ possible parabola.
A: 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. $\square$

