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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: y225-x29 = 1 - Mathematics and Statistics

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प्रश्न

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^2/9` = 1

बेरीज
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उत्तर

The equation of the hyperbola is `y^2/25 - x^2/9` = 1

Comparing with `y^2/"b"^2 - x^2/"a"^2` = 1, we get,

b2 = 25, a2 = 9

∴ b = 5, a = 3

(1) Length of transverse axis = 2b = 2(5) = 10

(2) Length of conjugate axis = 2a = 2(3) = 6

(3) Eccentricity = e = `sqrt("a"^2 + "b"^2)/"b"`

= `sqrt(25 + 9)/5`

= `sqrt(34)/5`

(4) be = `5(sqrt(34)/5) = sqrt(34)`

Coordinates of foci = (0, ± be) = `(0, ±sqrt(34))`

(5) `"b"/"e" = 5/((sqrt(34)/5)) = 25/sqrt(34)`

The equations of directrices are

y = `± "b"/"e"` i.e. y = `± 25/sqrt(34)`

(6) Length of latus rectum = `(2"a"^2)/"b"`

= `(2(9))/5`

= `18/5`

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Conic Sections - Hyperbola
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 7: Conic Sections - Exercise 7.3 [पृष्ठ १७४]

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बालभारती Mathematics and Statistics 1 (Arts and Science) [English] Standard 11 Maharashtra State Board
पाठ 7 Conic Sections
Exercise 7.3 | Q 1. (vii) | पृष्ठ १७४

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