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Find the Eccentricity, Coordinates of the Foci, Equation of Directrice and Length of the Latus-rectum of the Hyperbola . 3x2 − Y2 = 4 - Mathematics

Answer in Brief

Find the eccentricity, coordinates of the foci, equation of directrice and length of the latus-rectum of the hyperbola .

 3x2 − y2 = 4 

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Solution

 Equation of the hyperbola:  3x2 − y2 = 4

This can be rewritten in the following way:

\[\frac{3 x^2}{4} - \frac{y^2}{4} = 1\]

\[ \Rightarrow \frac{x^2}{\frac{4}{3}} - \frac{y^2}{4} = 1\]

This is the standard equation of a hyperbola, where 

\[a^2 = \frac{4}{3} \text { and }b^2 = 4\]

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

\[ \Rightarrow 4 = \frac{4}{3}( e^2 - 1)\]

\[ \Rightarrow e^2 - 1 = 3\]

\[ \Rightarrow e^2 = 4\]

\[ \Rightarrow e = 2\]

Coordinates of the foci are given by  \[\left( \pm ae, 0 \right)\], i.e.

\[\left( \pm \frac{4\sqrt{3}}{3}, 0 \right)\] . 

Equation of the directrices:

\[x = \pm \frac{a}{e}\]

\[x = \pm \frac{\sqrt{\frac{4}{3}}}{2}\]

\[ \Rightarrow \sqrt{3}x \pm 1 = 0\]

Length of the latus rectum of the hyperbola = \[\frac{2 b^2}{a}\] \[\Rightarrow \frac{2 \times 4}{\sqrt{\frac{4}{3}}} = 4\sqrt{3}\]

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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 27 Hyperbola
Exercise 27.1 | Q 3.4 | Page 13
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