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. $4x^2 -+3y^2 - 25 = 0$

Answer 1

Answer: (C)

We have $a = 3$ and $\frac{b^{2}}{a} = 4 \Rightarrow b^{2} = 12$
Hence, the equation of the hyperbola is
$\frac{x^{2}}{9}-\frac{y^{2}}{12} = 1$
$\Rightarrow 4x^{2} - 3y^{2} = 36 \Rightarrow 4x^{2} - 3y^{2} - 36 = 0$