What is the eccentricity of the ellipse x²/36 + y²/16 = 1 ?

To find the eccentricity of the ellipse given by the equation:

x²/36 + y²/16 = 1

Step 1: Identify the standard form of an ellipse
The general equation of an ellipse is:

x²/a² + y²/b² = 1

where:

"a" is the semi-major axis
"b" is the semi-minor axis
For the given equation, we compare it with the standard form:
a² = 36 → a = 6
b² = 16 → b = 4

Since a > b, the given ellipse is horizontal (major axis along the x-axis).

Step 2: Find the eccentricity formula
The eccentricity (e) of an ellipse is given by:

e = sqrt(1 - (b²/a²))

Substituting the values of a² and b²:

e = sqrt(1 - (16/36))
e = sqrt(1 - 4/9)
e = sqrt(5/9)
e = sqrt(5) / 3

Final Answer:
The eccentricity of the given ellipse is sqrt(5) / 3 or approximately 0.745.