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Find the equation of the hyperbola whose vertices are (± 6, 0) and one of the directrices is x = 4. - Mathematics

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Question

Find the equation of the hyperbola whose vertices are (± 6, 0) and one of the directrices is x = 4.

Sum
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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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Chapter 11: Conic Sections - Solved Examples [Page 196]

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NCERT Exemplar Mathematics [English] Class 11
Chapter 11 Conic Sections
Solved Examples | Q 10 | Page 196

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