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Question
The distance between the foci of a hyperbola is 16 and its eccentricity is `sqrt(2)`. Its equation is ______.
Options
x2 – y2 = 32
`x^2/4 - y^2/9` = 1
2x2 – 3y2 = 7
None of these
Solution
The distance between the foci of a hyperbola is 16 and its eccentricity is `sqrt(2)`. Its equation is x2 – y2 = 32.
Explanation:
We know that the distance between the foci = 2ae
∴ 2ae = 16
⇒ ae = 8
Given that e = `sqrt(2)`
∴ `sqrt(2)a` = 8
⇒ `a = 4sqrt(2)`
Now b2 = a2 (e2 – 1)
⇒ b2 = 32(2 – 1)
⇒ b2 = 32
So, the equation of the hyperbola is `x^2/a^2 - y^2/b^2` = 1
⇒ `x^2/32 - y^2/32` = 1
⇒ x2 – y2 = 32
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