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.
how far apart are the foci of an ellipse with a major axis of 34 feet and a minor axis of 16 feet
3 answers
AAAaannndd the bot gets it wrong yet again!
No, the distance between the foci is 2c = 30 ft
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.
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.