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1)

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


A) 4x23y2+36=0

B) 4x23y2+12=0

C) 4x23y236=0

D) 4x2+3y225=0

Answer:

Option C

Explanation:

We have a = 3 and b2a = 4

→ b2 = 12

Hence, the equation of the hyperbola is

x29y212=1

4x23y2=36 = 4x23y236=0