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Question
The equation of the hyperbola with vertices at (0, ± 6) and eccentricity `5/3` is ______ and its foci are ______.
Solution
The equation of the hyperbola with vertices at (0, ± 6) and eccentricity `5/3` is `y^2/36 - x^2/64` and its foci are (0, ± 10).
Explanation:
Let equation of the hyperbola is `- x^2/a^2 + y^2/b^2` = 1
Vertices are (0, ± b)
∴ b = 6 and e = `5/3`
We know that e = `sqrt(1 + a^2/b^2)`
⇒ `5/3 = sqrt(1 + a^2/36)`
⇒ `25/9 = 1 + a^2/36`
⇒ `a^2/36 = 25/9 - 1 = 16/9`
⇒ `a^2 = 16/9 xx 36`
⇒ `a^2` = 64
So the equation of the hyperbola is `(-x^2)/64 + y^2/36` = 1
⇒ `y^2/36 - x^2/64` = 1
And foci = (0, ± be) = `(0, +- 6 xx 5/3)` = (0, ± 10)
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