$$
\begin{aligned}
& \text { } 2 x^2+a y^2-8 x-2 a y+8-a=0.....(i) \\
& \Rightarrow\left(2 x^2-8 x\right)+\left(a y^2-2 a y\right)=a-8 \\
& \Rightarrow \frac{(x-2)^2}{a}+\frac{(y-1)^2}{2}=1
\end{aligned}
$$
We have $b^2=a^2\left(i-e^2\right)$ and given that major axis is parallel to $\mathrm{y}$-axis and $\mathrm{e}=\frac{\mathrm{1}}{\sqrt{3}}$
$$
\therefore a=2\left(1-\frac{1}{3}\right) \Rightarrow a=\frac{4}{3}
$$
Since slope of tangents $\frac{d y}{d x}=1$
Now, $4 x+2 a y \frac{d y}{d x}-8-2 a \frac{d y}{d x}=0$
Put $\frac{d y}{d x}=1, a=\frac{4}{3}$, we get
$$
3 x+2 y-8=0
$$
$y=\frac{8-3 x}{2}$ putting in (i), we get
$$
\begin{aligned}
& 15 x^2-60 x+52=0 \\
& \Rightarrow x=2 \pm \frac{4}{\sqrt{30}} \text { and } y=1 \pm \frac{6}{\sqrt{30}}
\end{aligned}
$$
$\therefore$ Equation of tangent is
$$
x-y-1 \pm \sqrt{\frac{10}{3}}=0
$$