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Question
Find the equation of the hyperbola with vertices at (0, ± 6) and e = `5/3`. Find its foci.
Solution
Since the vertices are on the y-axes (with origin at the mid-point)
The equation is of the form `y^2/a^2 - x^2/b^2` = 1.
As vertices are (0, ± 6)
a = 6
b2 = a2(e2 – 1)
= `36 25/9 - 1`
= 64
So the required equation of the hyperbola is `y^2/36 - x^2/64` = 1
And the foci are (0, ± ae) = (0, ± 10).
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