How far apart are the foci of an ellipse with a major axis of 34 feet and a minor axis of 16 feet?
The distance between the foci of an ellipse can be found using the formula:
c = √(a^2 - b^2)
Where a is the length of the semi-major axis and b is the length of the semi-minor axis. The distance between the foci is equal to two times c.
In this case, the length of the major axis is 34 feet, so the length of the semi-major axis is 17 feet. The length of the minor axis is 16 feet, so the length of the semi-minor axis is 8 feet.
Using the formula above, we can find the distance between the foci:
c = √(17^2 - 8^2) = √225 = 15
Therefore, the distance between the foci is twice c:
2c = 2(15) = 30 feet.
So, the foci are 30 feet apart.
To find the distance between the foci of an ellipse, we can use the equation:
c^2 = a^2 - b^2
Where:
- c is the distance between each focus and the center of the ellipse.
- a is the semi-major axis (half of the major axis).
- b is the semi-minor axis (half of the minor axis).
In this case, the major axis is 34 feet, so the semi-major axis (a) is 34 / 2 = 17 feet.
The minor axis is 16 feet, so the semi-minor axis (b) is 16 / 2 = 8 feet.
Now, we can calculate the distance between the foci (c):
c^2 = 17^2 - 8^2
c^2 = 289 - 64
c^2 = 225
To find c, we take the square root of both sides of the equation:
c = √225
c = 15 feet
Thus, the distance between the foci of the ellipse is 15 feet.