Consider the following lines
- $x-y-1=0$
- $x+y-5=0$
- $y=4$
The line 1 is the axis of the parabola, the line 2 is the tangent at the vertex to the same parabola, and the line 3 is another tangent to the same parabola at some point $P$.
Now let a circle $C$ circumscribe the triangle formed by tangent and normal at the point $P$ and the axis of the parabola.
Then how can I find the equation of the circle?
I found that the vertex of this parabola is $(3,2)$. Don't know how to proceed further.
HINT
I would say, help subtangent of parabola:
axis: $ x-y-1=0,\quad t_V:\,x+y-5=0, \quad t_P:\, y-4=0$
Theorem: Vertex bisects subtangent of parabola. -->
$P_1 = t_P \times \,$axis, $\quad\overline{P_1V}=\overline{VP_2}\Rightarrow P_2$
Perpendicular from P_2 to axis $\times\, t_P = P, \quad$ perpendicular from P to $t_P \times\,$axis $ = P_3$
$P_2$ = mid searched circle, = focus of parabola.