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
$\frac{x^2}{9}-\frac{y^2}{12}=1$
$\Rightarrow 4x^2-3y^2=36\Rightarrow 4x^2-3y^2-36=0$