11.Foci (0, ±13), the conjugate axis is of length 24.

The given coordinates of foci are ( 0,±13 ) and length of conjugate axis is 24.

Since the foci are on the y axis, the equation of the hyperbola is represented as,

y 2 a 2 − x 2 b 2 =1 , where y is the transverse axis.(1)

Since y axis is the transverse axis, coordinates of Foci = (0,±c)

∴c=13

Length of conjugate axis = 2b .

So, 2b=24 b=12

a 2 + b 2 = c 2 a 2 + 12 2 = 13 2 a 2 =169−144 a 2 =25 a=±5

Substitute the values of a and b in equation (1)

y 2 25 − x 2 144 =1

Thus, the equation of hyperbola with foci ( 0,±13 ) and length of conjugate axis 24 is y 2 25 − x 2 144 =1 .