Answer
$\displaystyle \frac{(x+4)^{2}}{16}+\frac{(y+5)^{2}}{25}=1$
New foci:$\quad (-4,-2)$ and $(-4,-8)$
New vertices:$\quad (-4,0)$ and $(-4, -10)$
New center:$\quad (-4,-5)$
Work Step by Step
Shift left by $4 \Rightarrow$ in the equation, replace $x$ with $x+4$
Shift down by $5 \Rightarrow$ in the equation, replace $y$ with $y+5$
New equation:$\displaystyle \quad \frac{(x+4)^{2}}{16}+\frac{(y+5)^{2}}{25}=1$
The ellipse $\displaystyle \frac{x^{2}}{16}+\frac{y^{2}}{25}=1$ has the form for which:
Foci on the y-axis: $\quad \displaystyle \frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1 \quad (a\gt b)$
$a=4,\quad b=5$
Center-to-focus distance: $c=\sqrt{a^{2}-b^{2}}=\sqrt{25-16}=3$
Foci:$\quad (0, \pm 3)$
Vertices:$\quad (0, \pm 5)$
Center:$\quad (0,0)$
The translations are such that $(x,y)\rightarrow(x',y')$, where
$\left\{\begin{array}{ll}
x'=x-4 & \text{... shift left}\\
y'=y-5 & \text{... shift down}
\end{array}\right.$
New foci:$\quad (0-4, \pm 3-5)\Rightarrow\quad (-4,-2)$ and $(-4,-8)$
New vertices:$\quad (0-4, \pm 5-5)\Rightarrow\quad (-4,0)$ and $(-4, -10)$
New center:$\quad (0-4,-5+0)\quad =(-4,-5)$
