Answer 1

Let the required equation of hyperbola be

`x^2/"a"^2 - y^2/"b"^2` = 1

Length of conjugate axis = 2b

Given, length of conjugate axis = 5

∴ 2b = 5

∴ b = `5/2`

∴ b² = `25/4`

Distance between foci = 2ae

Given, distance between foci = 13

∴ 2ae = 13

∴ ae = `13/2`

∴ a²e² = `169/4`

∴ Now, b² = a²(e² – 1)

∴ b² = a²e² – a

∴ `25/4 = 169/4` – a²

∴ a² = `169/4 - 25/4`

∴ a² =  `144/4` = 36

∴ The required equation of hyperbola is

`x^2/36 - y^2/(25/4)` = 1,

i.e., `x^2/36 - (4y^2)/25` = 1