Question

Answer the following:

Find the
(i) lengths of the principal axes
(ii) co-ordinates of the foci
(iii) equations of directrices
(iv) length of the latus rectum
(v) Distance between foci
(vi) distance between directrices of the curve

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

Answer 1

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

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

a² = 144 and b² = 25

∴ a = 12 and b = 5

i. Length of transverse axis = 2a = 2(12) = 24

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

∴ lengths of the principal axes are 24 and 10.

ii. b² = a²(e² – 1)

∴ 25 = 144 (e² – 1)

∴ `25/144` = e² – 1

∴ e² = `1 + 25/144`

∴ e² = `169/144`

∴ e = `13/12`   ...[∵ e > 1]

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

i.e., `"S"(12(13/12),0)` and `"S'"(-12(13/12),0)`

i.e., S(13, 0) and S'(–13, 0)

iii. Equations of the directrices are x = `± "a"/"e"`

i.e., x = `± 12/((13/12))`, i.e., x = `± 144/13`

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

= `(2(25))/12` 

= `25/6`

v. Distance between foci = 2ae = `2(12)(13/12)` = 26

vi. Distance between directrices = `(2"a")/"e"`

= `(2(12))/((13/12))`

= `288/13`