how far apart are the foci of an ellipse with a major axis of 34 feet and a minor axis of 16 feet
AAAaannndd the bot gets it wrong yet again!
No, the distance between the foci is 2c = 30 ft
I apologize for the mistake in my earlier response. You are correct. The distance between the two foci of the ellipse is given by the formula:
2c = 2√(a^2 - b^2)
Plugging in the values of the semi-major and semi-minor axes, we get:
2c = 2√(17^2 - 8^2) = 30 feet
Therefore, the distance between the foci of the ellipse with a major axis of 34 feet and a minor axis of 16 feet is 30 feet.
To find the distance between the foci of an ellipse, you can use the formula:
c = √(a^2 - b^2)
Where:
- c is the distance between the foci
- a is half of the major axis length
- b is half of the minor axis length
In this case, the major axis is 34 feet, so a = 34/2 = 17 feet.
Similarly, the minor axis is 16 feet, so b = 16/2 = 8 feet.
Plugging these values into the formula:
c = √(17^2 - 8^2)
c = √(289 - 64)
c = √225
c = 15 feet
Therefore, the distance between the foci of the given ellipse is 15 feet.
The distance between the two foci of an ellipse is given by 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. In this case, the major axis is 34 feet, so the semi-major axis is a = 17 feet. Similarly, the minor axis is 16 feet, so the semi-minor axis is b = 8 feet.
Plugging these values into the formula, we get:
c = √(17^2 - 8^2) = √225 = 15 feet
Therefore, the two foci of the ellipse are 15 feet apart.