Prove the following:

The intersection points of any three tangents to a parabola given by the formula $y(y-y_0)=2p(x-x_0)$ are vertices of a triangle whose orthocenter belongs to the directrix of the parabola and the circumcircle of the triangle passes through the focus of the parabola.

**My attempt:**

**Orthocenter problem part- edited** (old notes deleted to be less chaotic and I believe there is still some space for improvement of the first part):

In the meantime, I realized it would be better to just use the condition of tangency and plug some of its parameters into the formula for a line in the $xy$ plane.

Let $y=k_ix + l_i,\ i=1,2,3$ be a tangent line to a parabola. Then $$p=2k_il_i\implies l_i=\frac{\frac{p}2}{k_i}$$

Now, our equation becomes: $$\boxed{y=k_ix+\frac{\frac{p}2}{k_i}}$$ We can substitute $\frac{p}2$ by $\alpha$, so $$\boxed{y=k_ix+\frac{\alpha}{k_i}}$$ This way, our computations are getting easier. The equations of the three arbitrary tangents to a parabola are: $$\begin{aligned}y&=k_1x+\frac{\alpha}{k_1}\\y&=k_2x+\frac{\alpha}{k_2}\\y&=k_3x+\frac{\alpha}{k_3}\end{aligned}$$ The intersection point of the $i-$th and $j-$ th tangent line: $$k_ix+\frac{\alpha}{k_i}=k_jx+\frac{\alpha}{k_j}\implies x=\alpha\frac{\frac1{k_j}-\frac1{k_i}}{k_i-k_j}=\frac{\alpha}{k_ik_j}$$ $$y=k_i\cdot\frac{\alpha}{k_ik_j}+\frac{\alpha}{k_i}=\alpha\left(\frac1{k_i}+\frac1{k_j}\right)$$ $$\boxed{S_{ij}=\left(\frac{\alpha}{k_ik_j},\alpha\frac{k_i+k_j}{k_ik_j}\right)}$$

Now, we have to find the line perpendicular to the $k-$th tangent line passing through the point $S_{ij}$. $k_\perp=-\frac1{k_k}$ E.g., one altitude of the formed triangle belongs to the line: $$\begin{aligned}y-y_{S_{1,2}}&=k_\perp(x-x_{S_{1,2}})\iff &y&=-\frac1{k_3}x+\frac{\alpha}{k_1k_2k_3}+\alpha\left(\frac1{k_1}+\frac1{k_2}\right)\\&&y&=-\frac1{k_3}x+\alpha\left(\frac1{k_1}+\frac1{k_2}+\frac1{k_1k_2k_3}\right)\end{aligned}$$

So, the $x$ coordinate of the intersection of the three tangent lines: $$\begin{aligned}y&=-\frac1{k_3}x+\alpha\left(\frac1{k_1}+\frac1{k_2}+\frac1{k_1k_2k_3}\right)\\y&=-\frac1{k_2}x+\alpha\left(\frac1{k_1}+\frac1{k_3}+\frac1{k_1k_2k_3}\right)\\y&=-\frac1{k_1}x+\alpha\left(\frac1{k_2}+\frac1{k_3}+\frac1{k_1k_2k_3}\right)\end{aligned}$$ $$\begin{aligned}-\frac1{k_3}x+\alpha\left(\frac1{k_1}+\frac1{k_2}+\frac1{k_1k_2k_3}\right)&=-\frac1{k_2}x+\alpha\left(\frac1{k_1}+\frac1{k_3}+\frac1{k_1k_2k_3}\right)\\\left(\frac1{k_2}-\frac1{k_3}\right)x&=\left(\frac1{k_3}-\frac1{k_2}\right)\alpha\\x&=-\alpha=-\frac{p}2\end{aligned}$$

**End of the first part. The rest remains the same not to be off-topic**.

Since the points $P_1,P_2, P_3$ are close to the directrix, $\triangle ABC$ in my picture is obtuse and its orthocenter is outside the triangle, but it doesn't have to be so at all.

Let $A,B,C$ be the intersection points of the tangents. $P_2\in\overline{AC}$ due to $P_1\preceq P_2\preceq P_3$.

Let $A'\in\overline{BC}$ s.t. $AA'\perp BC$, $B'\in\overline{AC}$ s.t. $BB'\perp AC$ and $C'\in\overline{AB}$ s.t. $CC'\perp AB$.

The center $S$ of the circumscribed circle $q$ of $\triangle ABC$ is the intersection point of the bisectors $s_1,s_2,s_3$ of the sides $\overline{AB},\overline{BC}$ and $\overline{AC}$ respectively. Furthermore, $\underline{\text{each side bisector is parallel to one of the sides of the triangle}}$, i.e., $$s_1\parallel\overline{AA'}\ \&\ s_2\parallel\overline{BB'}\ \&\ s_3\parallel\overline{CC'} $$

If the orthocenter $T$ is an orthogonal projection of the point $P_2$ onto the directrix $x=-\frac{p}2$ of the parabola, and if the circumscribed circle $q$ really passes through the focus $$\boxed{F\left(\frac{p}2,0\right)\ \text{or}\ F\left(\frac{p}2+x_0,y_0\right)}$$ then $|TP_2|=|P_2F|$.

**zoomed:**
According to the notation in the picture:
$$\begin{aligned}\measuredangle AA'B&=\measuredangle A'CT=\measuredangle BFL\\\measuredangle B'TA&=\measuredangle ACA'=\measuredangle AFB\end{aligned}$$
I can see:
$$\triangle AB'T\sim\triangle BB'C\sim\triangle A'AC\sim AFB'$$
In particular: $\boxed{\triangle AB'T\cong\triangle AFB'\implies\ |TB'|=|B'F|\implies\triangle TFP_2\ \text{is isosceles}\ \implies |TP_2|=|P_2F|\ }$

Also:
we can prove **the orthogonal projections of the focus onto the three tangents belong to the tangent passing through the vertex of the parabola**, meaning those projections are collinear, which, by the *Simson theorem*, implies the focus belongs to the circumcircle of the triangle.

May I ask for advice on solving this task and improve the parts I might have done correctly to be concise as possible? Thank you in advance!

**P. S.** I found a related paper, but almost none of the information has been covered in our official literature.

Just in case, I found an answer by @JeanMarie recalling the fact the orthocenter of the observed triangle lies on the directrix of the parabola, but I couldn't think of proof.

**Update on a special case:**

I've also read **the directrix is the set of all the points in the plane we can draw two mutually perpendicular tangents from** (which can be proven via *Vieta's formulae* and the *condition of tangency*). In a right-triangle, the orthocenter is obviously in the vertex opposite to the hypothenuse.