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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: x225-y216 = 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:

`x^2/25 - y^2/16` = 1

योग
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उत्तर

The equation of the hyperbola is `x^2/25 - y^2/16` = 1

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

a2 = 25, b2 = 16

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

(2) Length of conjugate axis = 2b = 2(4) = 8

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

= `sqrt(25 + 16)/5`

= `sqrt(41)/5`

(4) ae = `5(sqrt(41)/5) = sqrt(41)`

Co-ordinates of foci ≡ (± ae, 0) = `(± sqrt(41), 0)`

(5) `"a"/"e" = 5/((sqrt(41)/5)) = 25/sqrt(41)`

The equations of directrices are

x = `± "a"/"e"` i.e., x = `± 25/sqrt(41)`

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

= `(2(16))/5`

= `32/5`

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Conic Sections - Hyperbola
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 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
अध्याय 7 Conic Sections
Exercise 7.3 | Q 1. (i) | पृष्ठ १७४

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