Answer:
[tex]h = -3\\k = 3[/tex]
[tex]a^2 = 25\\b^2 = 9[/tex]
[tex]\frac{(x+3)^2}{25} + \frac{(y-3)^2}{9} = 1[/tex]
Step-by-step explanation:
The general equation of an ellipse is as follows:
[tex]\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1[/tex]
Where the point (h, k) is the center of the circle
a is called semi major axis: horizontal distance from the ellipse to its center
b is the semi minor axis: vertical distance from the ellipse to its center.
See the attached image.
There you can notice that
[tex]a = 5\\\\a^2 = 25\\\\b = 3\\\\b^2=9[/tex]
The center is at point (-3, 3)
Thus:
[tex]h = -3\\k = 3[/tex]
Then the equation sought is:
[tex]\frac{(x-(-3))^2}{5^2} + \frac{(y-3)^2}{3^2} = 1[/tex]
Simplify
[tex]\frac{(x+3)^2}{5^2} + \frac{(y-3)^2}{3^2} = 1[/tex]
