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प्रश्न
Answer the following:
Find the equation of the hyperbola in the standard form if eccentricity is `3/2` and distance between foci is 12.
उत्तर
Let the required equation of hyperbola be `x^2/"a"^2 - y^2/"b"^2` = 1
Given, eccentricity (e) = `3/2`
Distance between foci = 2ae
Given, distance between foci = 12
∴ 2ae = 12
∴ `2"a"(3/2)` = 12
∴ 3a = 12
∴ a = `12/3` = 4
∴ a2 = 16
Now, b2 = a2(e2 – 1)
∴ b2 = `16[(3/2)^2 - 1]`
= `16(9/4 - 1)`
= `16(5/4)`
∴ b2 = 20
∴ The required equation of hyperbola is `x^2/16 - y^2/20` = 1.
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