<s>[INST] 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?
 [/INST] 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.
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