Find the foci of the ellipse 9x^2-18x+25y^2-100y-116=0
To find the foci of the ellipse, we first need to put the given ellipse equation in standard form:
9x^2 - 18x + 25y^2 - 100y - 116 = 0
9(x^2 - 2x) + 25(y^2 - 4y) = 116
9(x^2 - 2x + 1) + 25(y^2 - 4y + 4) = 116 + 9(1) + 25(4)
9(x-1)^2 + 25(y-2)^2 = 264
(x-1)^2 / (29.33)^2 + (y-2)^2 / (9.71)^2 = 1
Since the value under the square root in the x term is greater, the major axis is along x, i.e. it's a horizontal ellipse.
The formula for finding the foci of an ellipse with major axis a and minor axis b is:
c = sqrt(a^2 - b^2)
In our case:
a = 29.33
b = 9.71
c = sqrt(29.33^2 - 9.71^2) = sqrt(861.28 - 94.36) = sqrt(766.92) = 27.7
Since the ellipse is horizontal, the foci are located at (±c, 2).
Therefore, the foci of the ellipse are at (1 - 27.7, 2) and (1 + 27.7, 2) or about (-26.7, 2) and (28.7, 2).