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. None of these

Answer 1

The correct option is C $4x^2-3y^2-36=0$

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