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?
Best Answer
Given points $A=(1,1)$, $B=(0,-2)$, $T=(-1,-1)$ (which is the intersection point of the tangents), the parabola touching $AT$ at $A$ and $BT$ at $B$ can be constructed using this property of a parabola (see EDIT 2 at bottom for a proof):
Construct then the circle tangent to $AT$ at $T$ and passing through $B$, whose arc inside $\angle ATB$ is the locus of points at which $BT$ subtends an angle equal to the exterior angle. And similarly construct the circle tangent to $BT$ at $T$ and passing through $A$: the intersection of those circles, different from $T$, is the focus $F$ (see diagram below).
Finally, to construct the directrix, one can for example find the midpoint $M$ of $AB$ and subsequently the midpoint $P$ of $TM$, then $P$ is a point on the parabola and the directrix is the line perpendicular to $TM$ and intersecting $PT$ at a distance from $P$ equal to $PF$.
EDIT.
In this particular case things are even simpler, because tangents $AT$ and $BT$ are perpendicular: this implies that the directrix passes through $T$ and is perpendicular to line TM described above, while $AB$ is a focal chord (see here for a proof). Once the directrix is found, from points $A$ and $B$ it is easy to find focus $F$.
EDIT 2.
It may be useful to give a proof of the property mentioned at the beginning. Let then $F$ be the focus of the parabola, $A$, $B$ any two points on it, $H$, $K$ their projections on the directrix, $T$ the intersection point of the lines touching the parabola at $A$ and $B$ (see figure below).
As $AF=AH$ and $AT$ bisects $\angle FAH$, then $AT$ is the perpendicular bisector of $FH$ and $HT=FT$. Likewise, $FT=KT$ and $H$, $F$, $T$ belong to the same circle centred at $T$.
But $\alpha=\angle FAT=\angle HAT=\angle FHK$ (complementary of the same angle) and $\angle FHK=\angle FTB$ (inscribed angle and half of central angle subtending the same arc), hence $\angle FAT=\angle FTB=\alpha$. Similarly, $\angle FBT=\angle FTA=\beta$ and: $$ \angle AFT=\angle BFT=\pi-\alpha-\beta = \text{external angle formed by tangents $AT$ and $BT$.} $$