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The eccentricity of the hyperbola x2a2-y2b2 = 1 which passes through the points (3, 0) and (32,2) is ______. - Mathematics

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प्रश्न

The eccentricity of the hyperbola `x^2/a^2 - y^2/b^2` = 1 which passes through the points (3, 0) and `(3 sqrt(2), 2)` is ______.

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उत्तर

The eccentricity of the hyperbola `x^2/a^2 - y^2/b^2` = 1 which passes through the points (3, 0) and `(3 sqrt(2), 2)` is e2 = `13/9`.

Explanation:

Given that the hyperbola `x^2/a^2 - y^2/b^2` = 1 is passing through the points (3, 0) and `(3 sqrt(2), 2)`

So we get a2 = 9 and b2 = 4

Again, we know that b2 = a2(e2 – 1).

This gives 4 = 9(e2 – 1)

or e2 = `13/9`

or e = `sqrt(13)/3`.

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अध्याय 11: Conic Sections - Solved Examples [पृष्ठ २०१]

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एनसीईआरटी एक्झांप्लर Mathematics [English] Class 11
अध्याय 11 Conic Sections
Solved Examples | Q 23 | पृष्ठ २०१

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