Equation of the hyperbola with foci (0,± 5) and e = 53 is
yz9−yz16=1
xz16−yz9=1
xz12−yz13=1
Foci = ( 0 , ±5 ) = ( 0 , ± be)
b53 = 5 ⇒ b = 3
a2=9(259−1) = 16
x216−y29=−1
Find the equation of the hyperbola whose vertices are (0, ±3) and the foci are (0, ±5).
Find the equation of the hyperbola satisfying the given condition, Vertices (0,±3),foci(0,±5)