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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

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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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