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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:
3x2 – y2 = 4
Solution
The equation of the hyperbola is 3x2 – y2 = 4
i.e. `x^2/((4/3)) - y^2/4` = 1
Comparing with `x^2/"a"^2 - y^2/"b"^2` = 1, we get
a2 = `4/3`, b2 = 4
∴ a = `2/sqrt(3)`, b = 2
(1) Length of transverse axis = 2a = `2(2/sqrt(3)) = 4/sqrt(3)`
(2) Length of conjugate axis = 2b = 2(2) = 4
(3) Eccentricity = e = `sqrt("a"^2 + "b"^2)/"a"`
= `sqrt(4/3 + 4)/((2/sqrt(3))`
= 2
(4) ae = `(2/sqrt(3))(2) = 4/sqrt(3)`
Coordinates of foci = (± ae, 0) = `(± 4/sqrt(3), 0)`
(5) `"a"/"e" = ((2/sqrt(3)))/2 = 1/sqrt(3)`
The equations of directrices are
x = `± "a"/"e"` i.e., x = `± 1/sqrt(3)`
(6) Length of latus rectum = `(2"b"^2)/"a"`
= `(2(4))/((2/sqrt(3))`
= `4sqrt(3)`
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