Find the equation of the hyperbola whose foci are (0,12) and (0,-12) and length of the latus rectum is 36. Share with your friends Share 1 Manbar Singh answered this We have the coordinates of the foci of the given hyperbola are 0,±12.Since the coordinates of the foci of the given hyperbola are of the form of 0,±c, then it is a case of vertical hyperbola.Let the required equation of the hyperbola be,y2a2 - x2b2=1 .....1Clearly, we have c=12.Length of the latus rectum=36⇒2b2a=36⇒b2=18a .....2We know that,c2=a2+b2⇒122=a2+b2 ∵ c=12⇒a2+b2=144 .....3Substituting the value of b2 from 2 in 3, we geta2+18a-144=0⇒a2+24a-6a-144=0⇒aa+24-6a+24=0⇒a-6a+24=0⇒a-6=0 or a+24=0⇒a=6 or a=-24 rejected, as a can't be negativeSo, a=6Now, a2=62=36b2=18a=18×6=108Hence from 1, the required equation of the hyperbola isy236-x2108=1 2 View Full Answer
