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

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

3x2 − y2 = 4

#### 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