For the normal chord, $t_2=-t_1-\frac{2}{t_1}$ … $\left(1\right)$
Also, the chord subtends an angle of ${90}^{\mathrm{o}}$ at the vertex
$\therefore t_1{t}_2=-4$ … $\left(2\right)$

From $\left(1\right), t_1{t}_2=-t_1^2-2$ or $-4=t_1^2-2$

$\therefore t_1^2=2$
$\therefore t_2=-\frac{4}{t_1}=-\frac{4}{\sqrt{2}}=-2\sqrt{2}$ or $t_2^2=8$
Now, $P{Q}^2=a^2{\left(t_1^2-t_2^2\right)}^2+4a^2{\left(t_1-t_2\right)}^2$
$=1\cdot \left[{\left(2-8\right)}^2+4{\left(\sqrt{2}+2\sqrt{2}\right)}^2\right]=\left[36+72\right]$
$P{Q}^2=108$ or $PQ=6\sqrt{3}$ units