#### Answer

Ellipse:
- the vertices are $(3\pm \sqrt 8,4)$, $(3,4\pm \sqrt 6)$
- the foci are $(3\pm \sqrt{2},4)$
- the center is $(3,4)$

#### Work Step by Step

By completing the square, we can write the equation
$$
8 y^{2}+6 x^{2}-36 x-64 y+134=0
$$
in the form
$$
8 (y-4)^2+6 (x-3 )^2=48.
$$
Then we get
$$
\left(\frac{x-3}{ \sqrt 8}\right)^{2}+\left(\frac{y-4}{\sqrt 6}\right)^{2}=1
$$
which is an ellipse with $a=\sqrt 8, b=\sqrt 6$ and hence $ c=\sqrt{a^2-b^2}=\sqrt{2} $
So, we have:
- the vertices are $(3\pm \sqrt 8,4)$, $(3,4\pm \sqrt 6)$
- the foci are $(3\pm \sqrt{2},4)$
- the center is $(3,4)$