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