Question

The eccentricity of the hyperbola $\frac{(x-1)^{2}}{25}-\frac{(y+2)^{2}}{11}=1$ is:

• A. $\frac{5}{3}$
• B. $\frac{6}{5}$
• C. $\frac{7}{5}$
• D. $\frac{25}{11}$
• E. $\frac{5}{11}$

Hint:

For hyperbolas, $e > 1$. If your result is less than 1, you likely calculated eccentricity for an ellipse.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Eccentricity is a measure of how much a conic section deviates from being circular. For a hyperbola, it is always greater than 1. We need to calculate it from the standard equation.
Step 2: Key Formula or Approach:
The standard equation for a horizontal hyperbola is \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \).
The relationship between a, b, and the distance from the center to a focus (c) is given by:
\[ c^2 = a^2 + b^2 \] The eccentricity (e) is defined as:
\[ e = \frac{c}{a} \] Step 3: Detailed Explanation:
The given equation of the hyperbola is:
\[ \frac{(x-1)^2}{25} - \frac{(y+2)^2}{11} = 1 \] By comparing this to the standard form, we can identify \( a^2 \) and \( b^2 \).
The term under the positive part gives \( a^2 \):
\[ a^2 = 25 \implies a = 5 \] The term under the negative part gives \( b^2 \):
\[ b^2 = 11 \] Now, we find the value of c using the formula \( c^2 = a^2 + b^2 \):
\[ c^2 = 25 + 11 = 36 \] \[ c = \sqrt{36} = 6 \] Finally, we calculate the eccentricity using \( e = \frac{c}{a} \):
\[ e = \frac{6}{5} \] Step 4: Final Answer:
The eccentricity of the hyperbola is \( \frac{6}{5} \).