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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
Solution
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
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