Question

The eccentricity of the hyperbola \(\frac{x^2}{16} - \frac{y^2}{9} = 1\) is:

• A. \(5/4\)
• B. \(4/5\)
• C. \(3/4\)
• D. \(5/3\)

Hint:

For a standard hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the eccentricity is simply \(e = \frac{\sqrt{a^2 + b^2}}{a}\). This avoids dealing with fractions under the root.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question requires us to calculate the eccentricity of a given hyperbola in its standard form.

Step 2: Key Formula or Approach:
The standard equation of a horizontal hyperbola is:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \] For this standard hyperbola, the relationship between the semi-major axis (\(a\)), semi-minor axis (\(b\)), and eccentricity (\(e\)) is given by:
\[ b^2 = a^2(e^2 - 1) \] Which can be rearranged to find the eccentricity directly:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]

Step 3: Detailed Explanation:
The given equation of the hyperbola is:
\[ \frac{x^2}{16} - \frac{y^2}{9} = 1 \] Comparing this given equation with the standard form, we can identify the constants:
\[ a^2 = 16 \] \[ b^2 = 9 \] Now, substitute these values into the eccentricity formula:
\[ e = \sqrt{1 + \frac{9}{16}} \] To add the terms under the square root, find a common denominator:
\[ e = \sqrt{\frac{16}{16} + \frac{9}{16}} \] \[ e = \sqrt{\frac{16 + 9}{16}} \] \[ e = \sqrt{\frac{25}{16}} \] Taking the principal square root yields:
\[ e = \frac{5}{4} \]

Step 4: Final Answer:
The correct choice is (A).