Equation of the hyperbola with length of the latusrectum 4 and e = 3 is
2x2−16y2=1
3x2−15y2=1
16x2−2y2=1
8x2−y2=1
2b2a=4 e = 3
b2=2a
a2(e2−1)=2a
⇒ a(8) = 2 ⇒ a = 14 , b2=12
x2116−y212=1
⇒ 16 x2−2y2=1
Equation of the hyperbola with length of the latusrectum 92 and e = 54 is