Solution:
Given the graph of the ellipse as shown below:
The standard form of an ellipse whose major axis is parallel to thex-axis is
[tex]\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1[/tex]
The length of the major axis equals 2a.
Thus,
[tex]\begin{gathered} 2a=4-(-2) \\ \Rightarrow2a=6 \\ \text{divide both sides by 2} \\ \text{thus,} \\ a=\frac{6}{2} \\ \Rightarrow a=3 \end{gathered}[/tex]
We can determine the center (h,k) using the midpoint formula expressed as
[tex]\begin{gathered} (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}) \\ \Rightarrow(h,k)=(\frac{-2+4}{2},\frac{3+3}{2}) \\ \Rightarrow(h,k)=(1,3) \end{gathered}[/tex]
The co-vertices of the graphed ellipse is
[tex](h,k\pm b)[/tex]
This implies that
[tex]\begin{gathered} (h,\text{ k+b)}\Rightarrow(1,4) \\ (h,\text{ k-b)}\Rightarrow2 \end{gathered}[/tex]
Thus, when
[tex]\begin{gathered} k+b=4 \\ \text{where k=}3 \\ we\text{ have} \\ 3+b=4 \\ \Rightarrow b=4-3 \\ \therefore b=1 \end{gathered}[/tex]
Hence,
[tex]\begin{gathered} \text{the value for h is 1} \\ \text{the value }for\text{ k is }3 \\ \text{the value for }a\text{ is 3} \\ \text{the value for }b\text{ is 1} \end{gathered}[/tex]