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

41*67/2=1373.5

step-by-step explanation:

i just used a calculator