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Question
Equation of the hyperbola with eccentricty `3/2` and foci at (± 2, 0) is ______.
Options
`x^2/4 - y^2/5 = 4/9`
`x^2/9 - y^2/9 = 4/9`
`x^2/4 - y^2/9` = 1
None of these
Solution
Equation of the hyperbola with eccentricty `3/2` and foci at (± 2, 0) is `x^2/4 - y^2/5 = 4/9`.
Explanation:
Given that e = `3/2`
And foci = (± ae, 0) = (± 2, 0)
∴ ae = 2
`a xx 3/2` = 2
⇒ `a = 4/3`
Now we know that b2 = a2(e2 – 1)
b2 = `16/9(9/4 - 1)`
b2 = `16/9 xx 5/4`
b2 = `20/9`
So, the equation of the hyperbola is `x^2/(4/3)^2 - y^2/(20/9)` = 1
⇒ `(9x^2)/16 - (9y^2)/20` = 1
⇒ `x^2/16 - y^2/20 = 1/9`
⇒ `x^2/4 - y^2/5 = 4/9`
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