Find the center, vertices, and foci of ellipse. 4x^2+25y^2-50y=75
To find the center, vertices, and foci of the ellipse, we first need to put the equation in standard form:
First, we divide each term by 75 to get:
$x^2/18.75 + (y^2/3) - (2y/3) = 1$
Next, we complete the square for the y-terms by adding and subtracting 2^2 = 4:
$x^2/18.75 + (y^2/3) - (2y/3) + 1 = 1 + 4
This simplifies to:
$x^2/18.75 + (y^2/3) - (2y/3) + 1^2 = 5$
Now, divide everything by 5 to get the equation in standard form:
$\frac{x^2}{18.75} + \frac{(y-2)^2}{15}=1$
Given that the standard form of an ellipse is:
$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$
The center of the ellipse is (h,k), so the center of this ellipse is at (0,2).
The vertices of the ellipse are given by (h±a, k), so the vertices are at (±√18.75, 2).
The foci of the ellipse are given by (h±c, k), where c = √(a^2 - b^2), so we need to find c for our ellipse:
$b^2 = 15$
$a^2 = 18.75$
$c^2 = a^2 - b^2 = 18.75 - 15 = 3.75
$c = \sqrt{3.75} = 1.936$
Therefore, the foci of the ellipse are at (±1.936, 2).