Six, a circle center at the origin, the focus $\mathrm{F}$ is on the $\mathbf{x}$ axis, the line connecting this focus and the vertex $\mathrm{B}(0,2 \sqrt{2})$ has an inclination angle of $135^{\circ}$, find the coordinates of the intersection points of the line passing through $\mathrm{F}$ and perpendicular to $\mathrm{BF}$ with the ellipse.
Let the line passing through $\mathrm{F}$ and perpendicular to $\mathrm{BF}$ be 1.
$\because \mathrm{BF}$ has an inclination angle of $1.35^{\circ}, \therefore 1$ has an inclination angle of $45^{\circ}$.
$\because b=c=2 \sqrt{2}, \quad \therefore a=\sqrt{2} b=4$.
The ellipse equation is $\frac{x^{2}}{16}+\frac{y^{2}}{8}=1$.
The equation of the line is $\mathbf{x}-2 \sqrt{2}$.
Solving (1) and (2), substituting (9) into (1) gives
$$
8 x^{2}+16\left(x^{2}-4 \sqrt{2} x+8\right)=16 \times 8 \text {, }
$$

which simplifies to $24 x^{2}-64 \sqrt{2} x=0$.
Thus, $x_{1}=0, x_{2}=\frac{8 \sqrt{2}}{3}$.
Consequently, $\mathrm{y}_{1}=-2 \sqrt{2}, \mathrm{y}_{2}=\frac{2 \sqrt{2}}{3}$.
Therefore, the coordinates of the intersection points are $(0,-2 \sqrt{2})$ and
$$
\left(\frac{8 \sqrt{2}}{3}, \frac{2 \sqrt{2}}{3}\right) .
$$