Let equation of tangent be $y=mx$ (As it passes through $\left(0,0\right)$)

On solving with conic

$x^2-{\left(mx-1\right)}^2=2\Rightarrow x^2-\left(m^2{x}^2+1-2mx\right)=2$

$x^2\left(1-m^2\right)+2mx-3=0$

$D=0$

$\Rightarrow {\left(2m\right)}^2-4\left(1-m^2\right) \left(-3\right)=0$

$m^2+3\left(1-m^2\right)=0 \Rightarrow 2m^2=3\Rightarrow m=\frac{\sqrt{3}}{\sqrt{2}},\frac{-\sqrt{3}}{\sqrt{2}}$ (rejected, as tangent touches in first quadrant)

$\therefore$ Equation of tangent is

$y=\frac{\sqrt{3}}{\sqrt{2}}x$

For point of contact, solving with conic

$x^2-{\left[\frac{\sqrt{3}x}{\sqrt{2}}-1\right]}^2=2\Rightarrow 2x^2-{\left(\sqrt{3}x-\sqrt{2}\right)}^2=4$

$2x^2-\left(3x^2+2-2\sqrt{6}x\right)=4\Rightarrow -x^2+2\sqrt{6}x=6$

$x^2-2\sqrt{6}x=-6\Rightarrow {\left(x-\sqrt{6}\right)}^2=0\Rightarrow x=\sqrt{6}$

$\therefore a=\sqrt{6} \Rightarrow \left[9a\right]=\left[22.045\right]$

$=22$