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Answer:
[tex]\frac{(x-2)^2}{16}+\frac{(y-1)^2}{36}=1[/tex]
Step-by-step explanation:
Notice how the vertices are on the same x-coordinate and the endpoints of the minor axis are on the same y-coordinate. This indicates that the ellipse is vertical (meaning the ellipse has a vertical major axis of length [tex]2a[/tex]).
The equation for a vertical ellipse is [tex]\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1[/tex] where [tex](h,k)[/tex] is the center of the ellipse, [tex](h,k\pm a)[/tex] represents the coordinates of the vertices, and [tex](h\pm b,k)[/tex] represents the coordinates of the endpoints of the minor axis (also called co-vertices).
The value of [tex]h[/tex] is listed out for us as [tex]h=2[/tex].
The value of [tex]k[/tex] can be determined from taking the midpoint of the vertices. Because [tex]k=\frac{-5+7}{2}=1[/tex], then [tex]1-a=-5[/tex] and [tex]1+a=7[/tex] give us [tex]a=6[/tex].
Lastly, to figure out [tex]b[/tex], we solve the equations [tex]2+b=-2[/tex] and [tex]2-b=6[/tex] which both give [tex]b=-4[/tex] as the solution.
Now, plugging in all our values gives us [tex]\frac{(x-2)^2}{(-4)^2}+\frac{(y-1)^2}{6^2}=1[/tex] which translates to [tex]\frac{(x-2)^2}{16}+\frac{(y-1)^2}{36}=1[/tex].
Therefore, the final equation of the ellipse that satisfies the given conditions is [tex]\frac{(x-2)^2}{16}+\frac{(y-1)^2}{36}=1[/tex].
I've attached a graph of the ellipse with labels to help you visualize it.
