I am terrible at ellipse
The base of an auditorium is in the form of an ellipse 200ft long and 100 ft wide. A pin dropped near one focus can clearly be heard at the other focus. Determine the distance between the foci to the nearest tenth of a foot. Show each step.
You have
a = semi-major axis = 100
b = semi-minor axis = 50
a^2 = b^2+c^2
where c is the distance from the center to the focus.
Find c, and then the distance between the foci is 2c.
Well, it sounds like you really have an elliptical problem on your hands! But don't worry, I'm here to help and add a bit of humor to the mix.
Alright, let's get serious about this ellipse business. We know that the base of the auditorium is in the form of an ellipse, with dimensions of 200ft long and 100ft wide. Now, if you drop a pin near one focus, and it can be heard at the other focus, we need to figure out the distance between those two foci.
To find the distance between the foci of an ellipse, we need to use a handy formula. But hold on tight, because this formula might make you feel a bit elliptical:
The distance between the foci of an ellipse is given by the formula: c = sqrt(a^2 - b^2), where "c" is the distance between the foci, "a" is the length of the major axis, and "b" is the length of the minor axis.
In our case, we have a major axis of 200ft and a minor axis of 100ft. Let's plug those values into the formula:
c = sqrt(200^2 - 100^2)
Now let's calculate:
c = sqrt(40,000 - 10,000)
c = sqrt(30,000)
And the moment of truth... drumroll please... the distance between the foci is approximately 173.2 feet!
So there you have it! The distance between the foci of the ellipse-shaped auditorium is approximately 173.2 feet. Remember, math might not always be a laughing matter, but a little humor can always help ease the tension!
To determine the distance between the foci of the ellipse, we need to utilize the relationship between the length of the major axis (2a) and the distance between the foci (2c).
We are given that the base of the auditorium is in the form of an ellipse, with a length of 200ft (2a = 200ft) and a width of 100ft.
Step 1: Find the value of a.
The length of the major axis is given as 200ft, so we know that 2a = 200ft. We can find the value of a by dividing both sides of the equation by 2:
2a/2 = 200ft/2
a = 100ft
Step 2: Find the value of b.
The width of the ellipse is given as 100ft. The value of b represents the length of the minor axis, which is the distance from the center of the ellipse to the edge. In this case, b = 100ft.
Step 3: Find the value of c.
Using the relationship a^2 = b^2 + c^2, we can determine the value of c. Plug in the values of a and b into the equation:
100ft^2 = 100ft^2 + c^2
10000ft^2 = 10000ft^2 + c^2 - 10000ft^2
c^2 = 10000ft^2 - 10000ft^2
c^2 = 10000ft^2
To isolate c, we can take the square root of both sides of the equation:
c = √(10000ft^2)
c = 100ft
Step 4: Determine the distance between the foci.
The distance between the foci is given by 2c. Therefore, we multiply the value of c by 2 to find the distance between the foci:
2c = 2 * 100ft
2c = 200ft
The distance between the foci of the ellipse is 200ft.
To determine the distance between the foci of an ellipse, we need to know the lengths of the major and minor axes of the ellipse. In this case, the major axis is the length of the ellipse, which is 200 ft, and the minor axis is the width of the ellipse, which is 100 ft.
The formula to find the distance between the foci of an ellipse is given by:
c = √(a^2 - b^2),
where "c" is the distance between the foci, "a" is half the length of the major axis, and "b" is half the length of the minor axis.
Step 1: Find the values of "a" and "b."
Half the length of the major axis: a = 200/2 = 100 ft.
Half the length of the minor axis: b = 100/2 = 50 ft.
Step 2: Calculate the distance between the foci (c).
c = √(100^2 - 50^2)
c = √(10,000 - 2,500)
c = √7,500
Step 3: Round the result to the nearest tenth of a foot.
Using a calculator or the square root approximation, √7,500 ≈ 86.6025
Therefore, the distance between the foci of the given ellipse is approximately 86.6 ft.