Question

The eccentricity of the ellipse \(\frac{x^2}{36}+\frac{y^2}{16}=1\) is

• A. \(\frac{\sqrt{5}}{3}\)
• B. \(\frac{\sqrt{5}}{6}\)
• C. \(\frac{\sqrt{30}}{6}\)
• D. \(\frac{\sqrt{10}}{6}\)
• E. \(\frac{\sqrt{30}}{7}\)

Answer 1

The Correct Option is A

Step 1: Identify the standard form of the ellipse  
The given equation of the ellipse is: \[ \frac{x^2}{36} + \frac{y^2}{16} = 1 \] Comparing with the standard form: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] we get \( a^2 = 36 \) and \( b^2 = 16 \).

Step 2: Compute the eccentricity 
The eccentricity \( e \) of an ellipse is given by: \[ e = \frac{c}{a} \] where \( c \) is the focal distance, calculated as: \[ c^2 = a^2 - b^2 \] Substituting values: \[ c^2 = 36 - 16 = 20 \] \[ c = \sqrt{20} = 2\sqrt{5} \] Now, calculating \( e \): \[ e = \frac{c}{a} = \frac{2\sqrt{5}}{6} = \frac{\sqrt{5}}{3} \]

Final Answer: The eccentricity of the ellipse is \(\frac{\sqrt{5}}{3}\).