Question

The equation of the hyperbola with vertices at

\[ (0, \pm 6) \quad \text{and} \quad e = \frac{5}{3} \]

is:

• A. \( \frac{x^2}{36} - \frac{y^2}{64} = 1 \)
• B. \( \frac{y^2}{36} - \frac{x^2}{64} = 1 \)
• C. \( \frac{x^2}{64} - \frac{y^2}{36} = 1 \)
• D. \( \frac{y^2}{64} - \frac{x^2}{36} = 1 \)

Hint:

For a hyperbola, use the relationship between \( a \), \( b \), and the eccentricity to solve for the unknowns in the standard equation.

Answer 1

The Correct Option is B

Step 1: The standard equation of a hyperbola with vertical transverse axis is: \[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1, \] where \( a \) is the distance from the center to the vertices and \( b \) is related to the asymptotes. The eccentricity \( e \) is given by: \[ e = \frac{\sqrt{a^2 + b^2}}{a}. \] We are given that \( e = \frac{5}{3} \) and the vertices are at \( (0, \pm 6) \), so \( a = 6 \). 
Step 2: Using the formula for eccentricity: \[ \frac{5}{3} = \frac{\sqrt{a^2 + b^2}}{a} = \frac{\sqrt{36 + b^2}}{6}. \] Multiplying both sides by 6: \[ 10 = \sqrt{36 + b^2}. \] Squaring both sides: \[ 100 = 36 + b^2 \quad \Rightarrow \quad b^2 = 64 \quad \Rightarrow \quad b = 8. \] Thus, the equation of the hyperbola is: \[ \frac{y^2}{36} - \frac{x^2}{64} = 1. \]