1) The equation of the hyperbola with vertices (3 , 0), (-3 , 0) and semi-latus rectum 4 is given by : A) 4x2−3y2+36=0 B) 4x2−3y2+12=0 C) 4x2−3y2−36=0 D) 4x2+3y2−25=0 Answer: Option CExplanation:We have a = 3 and b2a = 4 → b2 = 12 Hence, the equation of the hyperbola is x29−y212=1 4x2−3y2=36 = 4x2−3y2−36=0