Solution:
The standard equation of an ellipse is expressed as
[tex]\begin{gathered} \frac{(x-h)^2}{a}+\frac{(y-k)^2}{b}=1\text{ } \\ \text{where} \\ (h,k)\text{ are the coordinate of the center of the ellipse.} \end{gathered}[/tex]
where
[tex]\begin{gathered} \text{the vertices}\Rightarrow(h\pm a,k),(h,k\pm b)_{} \\ \text{foci}\Rightarrow(h\pm c,k) \\ \text{where} \\ c^2=a^2-b^2 \end{gathered}[/tex]
Step 1:
Evaluate the center (h,k) along the vertices (-2,-1) and (8, -1)
Using the midpoint formula
[tex]\begin{gathered} (h,k)\text{ = (}\frac{-2+8}{2},\text{ }\frac{-1+(-1)}{2}) \\ =(\frac{6}{2},-\frac{2}{2}) \\ \Rightarrow(h,k)=(3,\text{ -1)} \end{gathered}[/tex]
Step 2:
From the length of the major axis of the ellipse given as 2a, we have
[tex]\begin{gathered} 2a=\sqrt[]{(}8-(-2))^2-(-1-(-1))^2 \\ 2a=10 \\ \Rightarrow a=5 \end{gathered}[/tex]
Step 3:
Recall that the foci is expressed as
[tex]\begin{gathered} \text{foci: }(h\pm c,k) \\ \text{thus, } \\ h+c=7,\text{ k=-1 or h-c=-1, k=-1} \\ where\text{ h=3} \\ \Rightarrow c=7-3=4 \end{gathered}[/tex]
but
[tex]\begin{gathered} c^2=a^2-b^2 \\ \Rightarrow16=25-b^2 \\ b^2=25-16=9 \\ \Rightarrow b=3 \end{gathered}[/tex]
Hence, the standard form of the ellipse becomes
[tex]\frac{(x-3)^2}{5}+\frac{(y+1)^2}{3}=1[/tex]
