SOLUTION
We want to find the major and minor axis of the ellipse
[tex]\frac{(x-3)^2}{12}+\frac{(y+4)^2}{24}=1[/tex]
Comparing the equation to the general equation of an ellipse given as
[tex]\frac{x^2}{a^2}+\frac{y^2}{b^2}=1[/tex]
we have
[tex]\begin{gathered} a^2=12,\text{ and } \\ b^2=24 \\ so\text{ } \\ a=\sqrt{12},b=\sqrt{24} \end{gathered}[/tex]
The major axis is the longest diameter of the ellipse, here it would be given as 2b, while the minor axis is the smaller diameter of the ellipse, here it would be given as 2a. So, we have
Major axis as
[tex]\begin{gathered} Major=2b \\ =2\times\sqrt{24} \\ =2\times\sqrt{4\times6} \\ =2\times2\times\sqrt{6} \\ 4\sqrt{6} \end{gathered}[/tex]
Hence the major axis is
[tex]4\sqrt{6}[/tex]
The minor axis becomes
[tex]\begin{gathered} 2a=2\times\sqrt{12} \\ 2\times\sqrt{4\times3} \\ 2\times2\times\sqrt{3} \\ =4\sqrt{3} \end{gathered}[/tex]
Hence the minor axis is
[tex]4\sqrt{3}[/tex]
