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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:
`y^2/25 - x^2/144` = 1
Solution
Given equation of the hyperbola is `y^2/25 - x^2/144` = 1
Comparing this equation with `y^2/"b"^2 - x^2/"a"^2` = 1, we get
b2 = 25 and a2 = 144
∴ b = 5 and a = 12
Length of transverse axis = 2b = 2(5) = 10
Length of conjugate axis = 2a = 2(12) = 24
Co-ordinates of vertices are
B(0, b) and B'(0, – b),
i.e., B(0, 5) and B'(0, – 5)
We know that
e = `sqrt("b"^2 + "a"^2)/"b"`
= `sqrt(25 + 144)/5`
= `sqrt(169)/5`
= `13/5`
Co-ordinates of foci are S(0, be) and S'(0, – be),
i.e., `"S"(0, 5(13/5))` and `"S'"(0, -5(13/5))`,
i.e., S(0, 13) and S'(0, – 13)
Equations of the directrices are y = `± "b"/"e"`.
∴ y = `± 5/((13/5))`
∴ y = `± 25/13`
Length of latus-rectum = `(2"a"^2)/"b"`
= `(2(144))/5`
= `288/5`
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