What is the center of the ellipse 4x^2+25y^2-50y=75
To find the center of an ellipse in the form (x-h)^2/a^2 + (y-k)^2/b^2 = 1, we first need to rearrange the given equation into this form.
4x^2 + 25y^2 - 50y = 75
4x^2 + 25(y^2 - 2y) = 75
4x^2 + 25(y^2 - 2y + 1) - 25 = 75
4x^2 + 25(y-1)^2 = 100
Now, we can rewrite this in the standard form:
x^2/(5)^2 + (y-1)^2/(2)^2 = 1
Comparing this to the standard form (x-h)^2/a^2 + (y-k)^2/b^2 = 1, we find that the center of this ellipse is at coordinates (h, k), which in this case is at (0, 1).
Therefore, the center of the ellipse 4x^2 + 25y^2 - 50y = 75 is at (0, 1).