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The Foci of the Hyperbola 2x2 − 3y2 = 5 Are - Mathematics

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प्रश्न

The foci of the hyperbola 2x2 − 3y2 = 5 are

विकल्प

  • \[( \pm 5/\sqrt{6}, 0)\]

  • (± 5/6, 0)

  • \[( \pm \sqrt{5}/6, 0)\]

  • none of these

MCQ
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उत्तर

\[( \pm 5/\sqrt{6}, 0)\]

The given equation of hyperbola is 

\[2 x^2 - 3 y^2 = 5\] It can be rewritten in the following way:

\[\frac{2 x^2}{5} - \frac{3 y^2}{5} = 1\]

\[ \Rightarrow \frac{x^2}{\frac{5}{2}} - \frac{y^2}{\frac{5}{3}} = 1\]

This is the standard equation of a parabola, where

\[a^2 = \frac{5}{2} \text { and }b^2 = \frac{5}{3}\].
The eccentricity can be calculated in the following way:

\[b^2 = a^2 \left( e^2 - 1 \right)\]

\[ \Rightarrow \frac{5}{3} = \frac{5}{2}\left( e^2 - 1 \right)\]

\[ \Rightarrow e^2 - 1 = \frac{2}{3}\]

\[ \Rightarrow e^2 = \frac{5}{3}\]

\[ \Rightarrow e = \sqrt{\frac{5}{3}}\]

Coordinates of the foci = \[\left( \pm ae, 0 \right) = \left( \pm \frac{5}{\sqrt{6}}, 0 \right)\]

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अध्याय 27: Hyperbola - Exercise 27.3 [पृष्ठ २०]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 27 Hyperbola
Exercise 27.3 | Q 17 | पृष्ठ २०

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