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If e and e' are the eccentricities of a hyperbola and its conjugate hyperbola respectively, prove that ee'1e2+1(e')2 = 1 - Mathematics and Statistics

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प्रश्न

If e and e' are the eccentricities of a hyperbola and its conjugate hyperbola respectively, prove that `1/"e"^2 + 1/("e""'")^2` = 1

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उत्तर

Let e be the eccentricity of the hyperbola `x^2/"a"^2 - y^2/"b"^2` = 1

∴ e = `sqrt("a"^2 + "b"^2)/"a"`

∴  e2 = `("a"^2 + "b"^2)/"a"^2`

Let e' be the eccentricity of the conjugate hyperbola `y^2/"b"^2 - x^2/"a"^2` = 1

∴ e' = `sqrt("b"^2 + "a"^2)/"b"`

∴ `"e'"^2` = `("a"^2 + "b"^2)/"b"^2`

∴ `1/"e"^2 + 1/"e'"^2 = "a"^2/("a"^2 + "b"^2) + "b"^2/("a"^2 + "b"^2)`

= `("a"^2 + "b"^2)/("a"^2 + "b"^2)`

= 1

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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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