
Let $A$ and $B$ be points on $x=1$ such that $|\overline{AB}| = 2$. With $R$ the point where $x=1$ meets the circle, define $\alpha := \angle ROA$ and $\beta := \angle ROB$. Let the "other" tangent from $A$ meet the circle at $S$; we see that $\angle ROS = 2\alpha$. Likewise, the "other" tangent from $B$ meets the circle at $T$ such that $\angle ROT = 2\beta$. Let these "other" tangents intersect at $P$, and note that $\overline{OP}$ bisects $\angle SOT$. Consequently, we can write
$$\begin{align}
P &= \sec(\alpha-\beta)\;\left( \cos(\alpha+\beta), \sin(\alpha+\beta) \right) \\
&= \left( \frac{\cos\alpha \cos\beta - \sin\alpha\sin\beta}{\cos\alpha\cos\beta + \sin\alpha \sin\beta}, \frac{\sin\alpha \cos\beta+\cos\alpha\sin\beta}{\cos\alpha\cos\beta + \sin\alpha \sin\beta} \right) \\
&=\left( \frac{1 - \tan\alpha\tan\beta}{1 + \tan\alpha \tan\beta}, \frac{\tan\alpha+\tan\beta}{1 + \tan\alpha \tan\beta} \right)
\end{align}$$
We can get the equation for the locus by eliminating $a := \tan\alpha$ and $b := \tan\beta$ from the system
$$x = \frac{1-a b}{1+ab} \qquad y = \frac{a+b}{1+ab}$$
subject to the "intercept condition"
$$a - b = 2$$
Without too much trouble, we arrive at the relation
$$ 2( 1 + x ) = y^2$$
which describes a parabola. $\square$