Given equation of ellipse is
$$
4 x^{2}+9 y^{2}=1
$$
On differentiating. we get $\Rightarrow \quad 8 x+18 yy'=0$
$\Rightarrow \quad y'=\frac{-8 x}{18 y}=m$
Also, $8 x=9 y \quad$ [equation of line]
On differentiating. we got $\Rightarrow \quad 8=9 y^{\prime}$
$\Rightarrow \quad y^{\prime}=\frac{8}{9}=m$
$[\because \text{tangents are parallel to this line}]$
So, by Eqs, we get $\frac{-8 x}{18 y}=\frac{8}{9}$
$\Rightarrow \quad-x=2 y$
$\Rightarrow \quad x=-2 y$
On substituting $x=-2 y$, we get
$$
\begin{array}{r}
4(-2 y)^{2}+9 y^{2}=1 \\
\Rightarrow \quad 16 y^{2}+9 y^{2}=1 \Rightarrow 25 y^{2}=1 \\
\Rightarrow \quad y=\pm \frac{1}{5} \quad \therefore \quad x=\mp \frac{2}{5}
\end{array}
$$
So, required points are $\left(\frac{2}{5},-\frac{1}{5}\right)$ and $\left(-\frac{2}{5}, \frac{1}{5}\right)$