X^2/49+y^2/36=1 How far from the center are the foci located?

The distance between the center and the foci is √13 ≅ 3.6 units

Step-by-step explanation:

* Lets revise the equation of the ellipse

- The standard form of the equation of an ellipse with center

(0, 0) and major axis parallel to the x-axis is

x²/a² + y²/b² = 1

# a > b

- The length of the major axis is 2a

- The coordinates of the vertices are (± a, 0)

- The length of the minor axis is 2b

- The coordinates of the co-vertices are (0, ± b)

- The coordinates of the foci are (± c, 0), where c² = a² - b²

* Lets solve the problem

∵ The equation is x²/49 + y²/36 = 1

∴ a² = 49

∴ b² = 36

∵ The coordinates of the foci are (± c, 0)

∵ c² = a² - b²

∴ c² = 49 - 36 = 13 ⇒ take square root for both sides

∴ c = ± √13

∴ The foci are (√13, 0) and (-√13, 0)

∵ The center is (0, 0)

∴ The distance between the center and the foci is c - 0 = c or

0 - (-c) = c

∴ The distance between the center and the foci = √13 - 0 = √13 units

* The distance between the center and the foci is √13 ≅ 3.6 units