The equation of the ellipse is required \frac{(x)^2}{16} +\frac{(y+4)^2}{25}=1
The required equation is
[tex]\frac{x^2}{16} +\frac{(y+4)^2}{25}=1[/tex]
It can be seen that the major axis is parallel to the y axis.
The major axis points are
[tex](h,k+a)=(0,1)\\(h,k-a)=(0,-9)\\k+a=1\\k-a=-9[/tex]
Subtracting the equations
[tex]2a=10\\\implies a=5[/tex]
The foci are
[tex](h,k+c)=(0,-1)\\(h,k-c)=(0,-7)[/tex]
[tex]k+c=-1\\k-c=-7[/tex]
Subtracting the equations
2c=6
c=3
k+c=-1 implies that k=-1-c
k=-1-3
k=-4
[tex]\frac{(x-h)^2}{b^2} +\frac{(x-h)^2}{a^2}=1[/tex]
So we get,
[tex]\frac{(x-0)^2}{4^2} +\frac{(y+4)^2}{5^2}=1[/tex]
[tex]\frac{(x)^2}{16} +\frac{(y+4)^2}{25}=1[/tex]
Therefore the equation is, [tex]\frac{(x)^2}{16} +\frac{(y+4)^2}{25}=1[/tex].
To learn more about the ellipse visit:
https://brainly.com/question/450229
