Question

The equation of the hyperbola with vertices \( (3, 0), (-3, 0) \) and semi-latus rectum 4 is given by:

• A. \( 4x^2 - 3y^2 + 36 = 0 \)
• B. \( 4x^2 - 3y^2 + 12 = 0 \)
• C. \( 4x^2 - 3y^2 - 36 = 0 \)
• D. \( 4x^2 - 3y^2 - 25 = 0 \)

Hint:

For a hyperbola with horizontal transverse axis, the equation takes the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), and the semi-latus rectum is used to find the value of \( b^2 \).

Answer 1

The Correct Option is C

Step 1: The equation of a hyperbola with the center at the origin and horizontal transverse axis is given by: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1. \] Step 2: For the given vertices, \( a = 3 \), and the semi-latus rectum gives \( b^2 = 12 \). Substituting these values into the equation, we get \( 4x^2 - 3y^2 - 36 = 0 \).

Final Answer: \[ \boxed{4x^2 - 3y^2 - 36 = 0} \]