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: Identify the curve.
The equation is $\dfrac{x^2}{16} - \dfrac{y^2}{9} = 1$, a horizontal hyperbola in standard form. We want its eccentricity $e$.
Step 2: Read off the constants.
Comparing with $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$, we see $a^2 = 16$ and $b^2 = 9$.
Step 3: Recall the eccentricity formula.
For a hyperbola, \[ e = \sqrt{1 + \frac{b^2}{a^2}} \] which always gives a value greater than $1$.
Step 4: Substitute the numbers.
\[ e = \sqrt{1 + \frac{9}{16}} \]
Step 5: Combine the fractions.
\[ e = \sqrt{\frac{16 + 9}{16}} = \sqrt{\frac{25}{16}} \]
Step 6: Take the square root.
\[ e = \frac{5}{4} \] Since $\frac{5}{4} > 1$, this is consistent with a hyperbola, and it is option (A).
\[ \boxed{e = \dfrac{5}{4}} \]