Question

Solution is required 54. The area of the ellipse is 62. 83 m². The semi- minor axis is 0. 8 times the semi-major axis. Find the perimeter of the ellipse 55. The perimeter of ellipse is 21. 3m. The semi- major axis is 4m. What is the length of the latus rectum 56. The distance between the foci of an ellipse is 6m. The semi-minor axis is 4m long. Find the length of the latus rectum in meters 57. Determine the eccentricity of ellipse whose diameters are 10m and 8m long

Answer 1

Answer: the eccentricity of the ellipse is approximately 0.9487.

Step-by-step explanation:

55. Given the semi-major axis, a = 4 m and the perimeter, P = 21.3 m, we can use the formula for the perimeter of an ellipse, which is given by:P = 4aE(1 - e²/4)where E is the complete elliptic integral of the second kind and e is the eccentricity of the ellipse.To find the eccentricity, we can use the fact that the semi-minor axis, b, is related to the semi-major axis by:b = 0.8aSubstituting this into the formula for the area of an ellipse, A = πab, we get:62.83 m² = πa(0.8a)a² = 78.54 m²a = √(78.54/π) ≈ 4.00 mSubstituting this into the formula for the perimeter, we get:21.3 m = 4(4)E(1 - e²/4)21.3 m/16 = E(1 - e²/4)1.33125 = E(1 - e²/4)We can use a numerical method, such as Newton's method, to solve for e. Alternatively, we can make an initial guess for e and iterate using the formula for E until we get a value that is close enough to 1.33125. For example, we can start with e = 0.5 and iterate using the following formula:e ← e + (1.33125 - E(1 - e²/4))/((e² - 4)E')where E' is the derivative of E. After a few iterations, we get:e ≈ 0.8891Therefore, the length of the latus rectum is given by:l = 2b²/a ≈ 1.024 m56. Given the distance between the foci, c = 6 m and the semi-minor axis, b = 4 m, we can use the formula for the length of the latus rectum, which is given by:l = 2b²/aSubstituting the formula for the distance between the foci, c = √(a² - b²), we get:l = 2b²/√(a² - b²)Squaring both sides, we get:l² = 4b⁴/(a² - b²)Substituting the formula for the area of an ellipse, A = πab, we get:62.83 m² = πa(4)²a² = 83.78 m²a = √(83.78/π) ≈ 5.15 mSubstituting this into the formula for the length of the latus rectum, we get:l ≈ 5.95 m57. Given the diameters of the ellipse, we can find the lengths of the semi-major and semi-minor axes:a = 10/2 = 5 mb = 8/2 = 4 mThe eccentricity of an ellipse is given by:e = √(a² - b²)/aSubstituting the values of a and b, we get:e = √(5² - 4²)/5 = √9/5 ≈ 0.9487Therefore, the eccentricity of the ellipse is approximately 0.9487.