Question

Find the length of transverse axis, length of conjugate axis, the eccentricity, the co-ordinates of foci, equations of directrices and the length of latus rectum of the hyperbola:

`y^2/25 - x^2/144` = 1

Answer 1

Given equation of the hyperbola is `y^2/25 - x^2/144` = 1

Comparing this equation with `y^2/"b"^2 - x^2/"a"^2` = 1, we get

b² = 25 and a² = 144

∴ b = 5 and a = 12

Length of transverse axis = 2b = 2(5) = 10

Length of conjugate axis = 2a = 2(12) = 24

Co-ordinates of vertices are

B(0, b) and B'(0, – b),

i.e., B(0, 5) and B'(0, – 5)

We know that

e = `sqrt("b"^2 + "a"^2)/"b"`

= `sqrt(25 + 144)/5`

= `sqrt(169)/5`

= `13/5`

Co-ordinates of foci are S(0, be) and S'(0, – be),

i.e., `"S"(0, 5(13/5))` and `"S'"(0, -5(13/5))`,

i.e., S(0, 13) and S'(0, – 13)

Equations of the directrices are y = `± "b"/"e"`.

∴ y = `± 5/((13/5))`

∴ y = `± 25/13`

Length of latus-rectum = `(2"a"^2)/"b"`

= `(2(144))/5`

= `288/5`