How far apart are the foci of an ellipse with a major axis of 26 feet and a minor axis of 24 feet?

10 ft

Got the whole test for ya right here, folks! For the Quadratic Relations and Conic Sections Test Part 1

1: A Ellipse; Domain {-5 < x < 5} Range {-2 < y < 2}

2: B x^2/4

3: C y = 1/10 x^2

4: C 9 inches

5: D x=1/12 (y - 4)^2 + 2

6: B (x - 4)^2 + (y - 5)^2 = 49

7: B (x + 8)^2 + (y + 4) = 25

8: C (x + 1)^2 + (y + 3)^2 = 4

9: B Center at (5, -3); Radius 5

10: B x^2/25 + y^2/169 = 1

11: D (0, +9)

12: C 10 feet

13: Graph C, the lines go ) (

14: D (0, +3)

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Well, let's see... the foci of an ellipse are like the eccentric relatives at a family reunion - they're always a bit distant. In this case, since the major axis is 26 feet and the minor axis is 24 feet, the distance between the foci can be calculated using a classic mathematical formula.

The formula is given by c = √(a^2 - b^2), where a is half the length of the major axis and b is half the length of the minor axis. So, let's plug in the values:

a = 26/2 = 13
b = 24/2 = 12

Using the formula, we have c = √(13^2 - 12^2), which simplifies to c ≈ √(169 - 144) ≈ √25 = 5.

Therefore, the foci of this eccentric ellipse are about 5 feet apart, just like those distant relatives at the family reunion who always seem worlds away.

To find the distance between the foci of an ellipse, we can use the formula:

c = √(a^2 - b^2)

where:
- c is the distance between the foci,
- a is the length of the semi-major axis (half of the major axis),
- b is the length of the semi-minor axis (half of the minor axis).

In this case:
- The length of the semi-major axis (a) is 26 feet / 2 = 13 feet,
- The length of the semi-minor axis (b) is 24 feet / 2 = 12 feet.

We can substitute these values into the formula:

c = √(13^2 - 12^2)
c = √(169 - 144)
c = √25
c = 5 feet

So, the foci of the ellipse are 5 feet apart.

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