Question

The length of transverse and conjugate axes in a hyperbola are 6 and 8, respectively. The eccentricity of the hyperbola, rounded off to TWO decimal places, is .......

Hint:

The eccentricity of a hyperbola is always greater than 1, and it can be calculated using the formula \( e = \sqrt{1 + \frac{b^2}{a^2}} \).

Answer 1

Correct Answer: 1.6 - 1.8

Step 1: Understanding the formula for eccentricity of a hyperbola.
The eccentricity \( e \) of a hyperbola is given by the formula: \[ e = \sqrt{1 + \frac{b^2}{a^2}}, \] where \( a \) is the length of the transverse axis and \( b \) is the length of the conjugate axis.

Step 2: Substituting the known values.
Given that the length of the transverse axis \( a = 6 \) and the length of the conjugate axis \( b = 8 \), we substitute these values into the formula: \[ e = \sqrt{1 + \frac{8^2}{6^2}} = \sqrt{1 + \frac{64}{36}} = \sqrt{1 + 1.7778} = \sqrt{2.7778}. \]

Step 3: Calculating the eccentricity.
\[ e = \sqrt{2.7778} \approx 1.67. \]

Step 4: Conclusion.
The eccentricity of the hyperbola is \( \boxed{1.67} \).