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Question
Find the equation of the hyperbola referred to its principal axes:
whose distance between foci is 10 and eccentricity `5/2`
Solution
Let the required equation of hyperbola be `x^2/"a"^2 - y^2/"b"^2` = 1.
Given, eccentricity (e) = `5/2`
Distance between foci = 2ae
Given, distance between foci = 10
∴ 2ae = 10
∴ ae = `10/2` = 5
∴ `"a"(5/2)` = 5
∴ a = 2
∴ a2 = 4
Now, b2 = a2(e2 – 1)
∴ b2 = `4[(5/2)^2 - 1]`
= `4(25/4 - 1)`
= `4(21/4)`
∴ b2 = 21
∴ The required equation of hyperbola is `x^2/4 - y^2/21` = 1.
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