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Question
Find the equation of the hyperbola with eccentricity `3/2` and foci at (± 2, 0).
Solution
Given that e = `3/2` and foci at (± 2, 0)
We know that foci = (± ae, 0)
∴ ae = 2
⇒ `a xx 3/2` = 2
⇒ `a = 4/3`
⇒ `a^2 = 16/9`
We know that b2 = a2(e2 – 1)
⇒ `b^2 = 16/9(9/4 - 1)`
= `16/9 xx 5/4`
= `20/9`
So, the equation of the hyperbola is `x^2/(16/9) - y^2/(20/9)` = 1
⇒ `(9x^2)/16 - (9y^2)/20` = 1
⇒ `x^2/4 - y^2/5 = 4/9`
Hence, the required equation is `x^2/4 - y^2/5 = 4/9`.
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