<s>[INST] A parabola touches the bisectors of the angles formed by lines $x+2y+3=0$ and $2x+y+3=0$ at $(1,1)$ and $(0,-2)$. Find its focus and directrix. 
A parabola touches the bisectors of the angles obtained by the lines $x+2y+3=0$ and $2x+y+3=0$ at the points $(1,1)$ and $(0,-2)$. Then find its focus and the equation of the directrix.

My approach is as follows:
The equation of bisector is 
$$\frac{x+2y+3}{\sqrt{5}}= \pm \frac{2x+y+3}{\sqrt{5}}$$
We get the required bisectors as $x-y=0$ and $3x+3y+6=0$, or $x+y+2=0$.
$x-y=0$ is tangent to the parabola at $(1,1)$, whereas $x+y+2=0$ is tangent to the parabola at $(0,-2)$. 
From here, how do I proceed?
 [/INST] Since angular bisectors are perpendicular to each other and are tangent to the parabola at $A(1,1)$ and $B(0,2)$, the point of intersection of the two given lines $(-1,-1)$ is a point on the directrix of the parabola.
The median through $P$ to the Archimedes triangle $PAB$ is known to be parallel to axis, which gives the slope of directrix as $-3$. So foot of perpendicular from $A$ on to directrix is $\left(-\dfrac{7}{5},\dfrac{1}{5}\right)$. This is also the image of the focus in the tangent $PA \equiv x-y=0$ which is easily found as $\boxed{\left(\dfrac{1}{5},-\dfrac{7}{5}\right)}$
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