Here a proof of a generalization of your statement in term of synthetic projective geometry.
Let $\Gamma$ be a conic in a plane $\alpha$ and let $\Pi$ be a
projectivity of $\Gamma$ onto itself with axis $p$ and center $P$ not
contained in $p$. Then the locus of points of the intersections of the
$\Pi$-corrispondent tangents of $\Gamma$ is a conic $\Psi$.
Moreover $\Gamma$ and $\Psi$ induces the same involution on $p$.
We identify a conic with its associated polarity in the plane.
Recall that a conic $\Gamma$ induces on a line $p$, not tangent to it, an involution which maps each point of p into the intersection of its polar with $p$.
First note that $p$ is not tangent to $\Gamma$ because it not contains its pole.
Let $D,E$ be the intersections (real or immaginary) of $p$ with $\Gamma$.
The projectivity $\Pi$ is determined by the image of a tangent $u$ to $\Gamma$.
Thus let $v=\Pi(u)$ and $A=u\cap v$.
If $U=\Gamma(u)$ and $V=\Gamma(v)$ are the respective contact points, then $\Pi(U)=V$. Moreover, $D$ and $E$ are fixed point of $\Pi$.
Let $\Delta$ be the (only) homography of the plane having $D,E,P$ as fixed points
and mapping $U$ in $A$.
We claim that $\Delta$ maps $\Gamma$ onto a conic $\Psi$ which is the locus of the points of the form $t\cap\Pi(t)$ for $t$ tangent to $\Gamma$.
We first show that for each point $T$ of $\Gamma$ we have $\Delta(T)\subset t$, because $t=\Gamma(T)$ is the line tangent to $\Gamma$ at $T$.
Since $\Delta$ fixes $D$, it induces a projectivity of the pencil of the lines at $D$ into itself.
In particular, for each line $r$ containing $D$, $\Delta(r)$ is a line containing $D$ and $\Delta$ induces a perspectivity of $r$ onto $\Delta(r)$ whose center $C$ belongs to the line $e=PE$ which is tangent to $\Gamma$ and fixed by $\Delta$.
Similarly, $\Delta$ induces a perspectivity from $PU$ to $PA$ with center $J=p\cap u$.
Consequently, $\Delta(r\cap PU)=(r\cap PU)J\cap PA$ and $C=((r\cap PU)J\cap PA)(r\cap PU)\cap e=(r\cap PU)J\cap e$.
If $R=\Gamma(r)$, then $CR$ is tangent to $\Gamma$ by Steiner's theorem.
This proves the assertion.
By a similar argument, also $\Delta(T)\subset\Pi(t)$ for each point $T$ of $\Gamma$. Consequently, we get $\Delta(T)=t\cap\Pi(t)$ for each point $T$ of $\Gamma$ where $t$ denote the tangent to $\Gamma$ at $T$.
It remains to show that $\Gamma$ and $\Psi$ induces the same involution on $p$.
Let $X$ be a point of $p$ and $X'$ be its conjugate respect to $\Gamma$.
Then $DEXX'$ is an harmonic group, thus $DE\Delta(X)\Delta(X')$ is an harmonic group as well. But $\Delta(X)$ and $\Delta(X')$ are points of $p$ conjugated respect to $\Psi$ and this concludes the proof.
Finally we deduce your statement from this.
Let $\Gamma$ be a parabola, $p$ be its directrix and $P$ it focus.
A non-involutory projectivity $\Pi$ of $\Gamma$ into itself with center $P$ and axis $p$ induces on the line at infinity a non-involutory projectivity which commutes with the absolute involution, that's a direct congruence; consequently, $\Pi$ rotate each tangent line by the same angle.
Then $\Pi$ maps $\Gamma$ into a conic $\Psi$ which induces on $p$ (and $P$) the same involution as $\Gamma$, that's the right-angle involution.
This proves that $\Psi$ has focus $P$ and directix $p$ as well.
In order to show that $\Psi$ is an hyperbola, it's enough to prove that the line at infinity $w$ is secant $\Psi$. But this follows at once by noting that the point at infinity of the line $\Pi(w)$ belongs to $\Psi$ and it's no a contact point.