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प्रश्न
Find the equation of the hyperbola referred to its principal axes:
whose distance between directrices is `8/3` and eccentricity is `3/2`
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उत्तर
Let the required equation of hyperbola be `x^2/"a"^2 - y^2/"b"^2` = 1
Given, eccentricity (e) = `3/2`
Distance between directrices = `(2"a")/"e"`
Given, distance between directrices = `8/3`
∴ `(2"a")/"e" = 8/3`
∴ `(2"a")/(3/2) = 8/3`
∴ `(4"a")/3 = 8/3`
∴ a = 2
∴ a2 = 4
∴ Now, b2 = a2(e2 – 1)
∴ b2 = `4[(3/2)^2 - 1]`
= `4(9/4 - 1)`
= `4(5/4)`
∴ b2 = 5
∴ The required equation of hyperbola is `x^2/4 - y^2/5` = 1.
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