The equation of the hyperbola whose centre is (5, 2), vertex is (9, 2) and the length of conjugate axis is 6 is
(x−5)216−(y−2)29=−1
(x−2)25−(y−2)29=1
(x+2)27−(y+2)29=1
(x−2)217+(y−2)219=1
CA=a⇒a=4 and b=3 ∴Hyperbola is(x−5)216−(y−2)29=1
Find the equation of the hyperbola whose foci are at (0,±6) and the length of whose conjugate axis is 2√11.