Answer
$ \displaystyle \frac{(x+2)^{2}}{6}+\frac{(y+1)^{2}}{9}=1$
New foci:$\quad (-2, -1\pm\sqrt{3})$
New vertices:$\quad (-2,2)$ and $(-2,-4)$
New center:$\quad (-2,-1)$
Work Step by Step
Shift left by $2 \Rightarrow$ in the equation, replace $x$ with $x+2$
Shift down by $1 \Rightarrow$ in the equation, replace $y$ with $y+1$
New equation:$\displaystyle \quad \frac{(x+2)^{2}}{6}+\frac{(y+1)^{2}}{9}=1$
The ellipse $\displaystyle \frac{x^{2}}{6}+\frac{y^{2}}{9}=1$
Foci on the y-axis: $\quad \displaystyle \frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1 \quad (a\gt b)$
$a=3,\quad b=\sqrt{6}$
Center-to-focus distance: $c=\sqrt{a^{2}-b^{2}}=\sqrt{9-3}=\sqrt{3}$
Foci:$\quad (0, \pm\sqrt{3})$
Vertices:$\quad (0, \pm 3)$
Center:$\quad (0,0)$
The translations are such that $(x,y)\rightarrow(x',y')$, where
$\left\{\begin{array}{ll}
x'=x-2 & \text{... shift left}\\
y'=y-1 & \text{... shift down}
\end{array}\right.$
New foci:
$\quad (0-2, -1\pm\sqrt{3})\Rightarrow\quad (-2, -1\pm\sqrt{3})$
New vertices:
$\quad (0-2,-1 \pm 3)\Rightarrow\quad (-2,2)$ and $(-2,-4)$
New center:
$\quad (0-2,-1+0)\quad =(-2,-1)$
