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^2/100 - y^2/25` = + 1

Answer 1

Given equation of the hyperbola is `x^2/100 - y^2/25` = + 1

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

a² = 100 and b² = 25

∴ a = 10 and b = 5

(1) Length of transverse axis = 2a = 2(10) = 20

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

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

= `sqrt(100 + 25)/10`

= `sqrt(125)/10`

= `(5sqrt(5))/10`

= `sqrt(5)/2`

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

i.e., `"S"(10(sqrt5/2),0)` and `"S'"(-10(sqrt5/2),0)`,

i.e., `"S"(5sqrt5,0)` and `"S'"(-5sqrt5,0)`

(5) Equations of the directrices are x = `± "a"/"e"`

∴ x = `± 10/((sqrt5/2))`

∴ x = `± 20/sqrt5`

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

= `(2(25))/10`

= 5