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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^2/25 - y^2/16` = – 1
Solution
Given equation of the hyperbola is `x^2/25 - y^2/16` = – 1.
∴ `y^2/16 - x^2/25` = 1
Comparing this equation with
`y^2/"b"^2 - x^2/"a"^2` = 1, we get
b2 = 16 and a2 = 25
∴ b = 4 and a = 5
Length of transverse axis = 2b = 2(4) = 8
Length of conjugate axis = 2a = 2(5) = 10
Co-ordinates of vertices are
B(0, b) and B'(0, – b)
i.e., B(0, 4) and B'(0, – 4)
We know that
e = `sqrt("b"^2 + "a"^2)/"b"`
= `sqrt(16 + 25)/4`
= `sqrt(41)/4`
Co-ordinates of foci are S(0, be) and S'(0, –be),
i.e., `"S"(0, 4(sqrt(41)/4))` and `"S""'"(0, - 4(sqrt(41)/4))`,
i.e., `"S"(0, sqrt(41))` and `"S""'"(0, -sqrt(41))`
Equations of the directrices are y = `± "b"/"e"`.
∴ y = `± 4/(sqrt(41)/4`
∴ y = `± 16/sqrt(41)`
Length of latus-rectum = `(2"a"^2)/"b"`
= `(2(25))/4`
= `25/2`.
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