Let $P(h,k)$ be the point from which tangents are drawn.
So, combined equation of $PA$ and $PB$ is
${\left(xh+yk-1\right)}^2=\left(x^2+y^2-1\right)\left(h^2+k^2-1\right)$ $\left(\because T^2=S{S}_1\right)$
Putting $x=1,$ we get,

 ${\left(h-1\right)}^2+2ky\left(h-1\right)=y^2\left(h^2-1\right)$
$\Rightarrow \left(h+1\right)y^2-2ky-\left(h-1\right)=0$
where roots are $y_1 \& y_2$ and $\left|y_1-y_2\right|=2$
$\Rightarrow \frac{\sqrt{{\left(2k\right)}^2+4\left(h+1\right)\left(h-1\right)}}{\left|h+1\right|}=2$
$\Rightarrow y^2+x^2-1=x^2+2x+1$
$\Rightarrow y^2=2\left(x+1\right)$