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Question
Find the equation of the hyperbola whose vertices are (± 6, 0) and one of the directrices is x = 4.
Solution
As the vertices are on the x-axis and their middle point is the origin
The equation is of the type `x^2/a^2 - y^2/b^2` = 1.
Here b2 = a2(e2 – 1)
Vertices are (± a, 0)
And Directrices are given by x = `+- a/e`.
Thus `a= 6, a/e` = 4
And so e = `3/2`
Which gives b2 = `36 9/4 - 1` = 45
Consequently, the required equation of the hyperbola is `x^2/36 - y^2/45` = 1
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