what is foci of (x-2)^2/16 + (y+3)^2/36
The given equation is in the standard form of an ellipse:
(x-h)^2/a^2 + (y-k)^2/b^2 = 1
Comparing this with the given equation:
(x-2)^2/16 + (y+3)^2/36 = 1
We can see that h = 2, k = -3. The center of the ellipse is at the point (h,k), so the center is (2, -3).
Now, let's find the values of a and b to determine the foci.
The distance from the center to each focus is given by c, where c^2 = a^2 - b^2.
In this case, a^2 = 36 and b^2 = 16, so c^2 = 36 - 16 = 20.
Taking the square root of both sides, we find c = √20 = 2√5.
The foci are located (h±c, k). Plugging in the values, we have two foci:
Foci 1: (2+2√5, -3)
Foci 2: (2-2√5, -3)
So, the foci of the given ellipse are (2+2√5, -3) and (2-2√5, -3).