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:

`x^2/25 - y^2/16` = – 1

Answer 1

Given equation of the hyperbola is `x^2/25 - y^2/16` = – 1.

∴ `y^2/16 - x^2/25` = 1

Comparing this equation with

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

b² = 16 and a² = 25

∴ b = 4 and a = 5

Length of transverse axis = 2b = 2(4) = 8

Length of conjugate axis = 2a = 2(5) = 10

Co-ordinates of vertices are

B(0, b) and B'(0, – b)

i.e., B(0, 4) and B'(0, – 4)

We know that

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

= `sqrt(16 + 25)/4`

= `sqrt(41)/4`

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

i.e., `"S"(0, 4(sqrt(41)/4))` and `"S""'"(0, - 4(sqrt(41)/4))`,

i.e., `"S"(0, sqrt(41))` and `"S""'"(0, -sqrt(41))`

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

∴ y = `± 4/(sqrt(41)/4`

∴ y = `± 16/sqrt(41)`

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

= `(2(25))/4`

= `25/2`.