# 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: x2100-y225 = + 1 - Mathematics and Statistics

Sum

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

#### Solution

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

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

a2 = 100, 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) ae = 5(sqrt(5)/2) = sqrt(10)

Coordinates of foci= (± ae, 0) = (± sqrt(5), 0)

(5) "a"/"e" = 5/((sqrt(5)/2)) = 20/sqrt(5)

The equations of directrices are

x = ± "a"/"e" i.e. x = ± 20/sqrt(5)

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

= (2(25))/10

= 5.

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