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Question
Answer the following:
Find the
(i) lengths of the principal axes
(ii) co-ordinates of the foci
(iii) equations of directrices
(iv) length of the latus rectum
(v) Distance between foci
(vi) distance between directrices of the curve
`x^2/144 - y^2/25` = 1
Solution
Given equation of the hyperbola `x^2/144 - y^2/25` = 1
Comparing this equation with `x^2/"a"^2 - y^2/"b"^2` = 1
a2 = 144 and b2 = 25
∴ a = 12 and b = 5
i. Length of transverse axis = 2a = 2(12) = 24
Length of conjugate axis = 2b = 2(5) = 10
∴ lengths of the principal axes are 24 and 10.
ii. b2 = a2(e2 – 1)
∴ 25 = 144 (e2 – 1)
∴ `25/144` = e2 – 1
∴ e2 = `1 + 25/144`
∴ e2 = `169/144`
∴ e = `13/12` ...[∵ e > 1]
Co-ordinates of foci are S(ae, 0) and S'(–ae, 0)
i.e., `"S"(12(13/12),0)` and `"S'"(-12(13/12),0)`
i.e., S(13, 0) and S'(–13, 0)
iii. Equations of the directrices are x = `± "a"/"e"`
i.e., x = `± 12/((13/12))`, i.e., x = `± 144/13`
iv. Length of latus rectum = `(2"b"^2)/"a"`
= `(2(25))/12`
= `25/6`
v. Distance between foci = 2ae = `2(12)(13/12)` = 26
vi. Distance between directrices = `(2"a")/"e"`
= `(2(12))/((13/12))`
= `288/13`
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