what is the foci of the ellipse
X2 over 81 + Y2 over 49=1
the ellipse would be horizontal (since a > b), so the foci would be (h +/- c, k). h is 0 and k is also 0. a is 9 and b is 7. a^2 = b^2 + c^2
81 = 49 + c^2
32 = c^2
4 x the square root of 2 = c
so the foci would be (+/- the square root of 2, 0)
To find the foci of an ellipse, you need to use the equation of the ellipse and its properties. The general equation for an ellipse with its center at the origin is:
(x^2 / a^2) + (y^2 / b^2) = 1
In this equation, a represents the semi-major axis (half of the length of the major axis) and b represents the semi-minor axis (half of the length of the minor axis).
In your specific equation: (x^2 / 81) + (y^2 / 49) = 1
By comparing it to the general equation, we can see that a^2 = 81 and b^2 = 49. Therefore, a = 9 and b = 7.
The foci of an ellipse are given by the formula c^2 = a^2 - b^2, where c represents the distance from the center to each focus. In this case, we have:
c^2 = 9^2 - 7^2
c^2 = 81 - 49
c^2 = 32
To find the values of c, we take the square root of both sides:
c = √32
c ≈ 5.66
Therefore, the foci of the given ellipse are located at a distance of approximately 5.66 units from the center, along the major axis.