Find the Eccentricity, Coordinates of the Foci, Equation of Directrice and Length of the Latus-rectum of the Hyperbola . 2x2 − 3y2 = 5. - Mathematics

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

2x2 − 3y2 = 5.

Solution

Equation of the hyperbola: 2x2 − 3y2 = 5.

This can be rewritten in the following manner:

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

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

This is the standard equation of a hyperbola, where

$a^2 = \frac{5}{2} \text { and }b^2 = \frac{5}{3}$ .

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

$\Rightarrow \frac{5}{3} = \frac{5}{2}( e^2 - 1)$

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

$\Rightarrow e^2 = \frac{5}{3}$

$\Rightarrow e = \sqrt{\frac{5}{3}}$

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

$\left( \pm \frac{5\sqrt{6}}{6}, 0 \right)$.

Equation of the directrices: $x = \pm \frac{a}{e}$

$x = \pm \frac{\sqrt{\frac{5}{2}}}{\sqrt{\frac{5}{3}}}$

$\Rightarrow x = \pm \frac{\sqrt{3}}{\sqrt{2}}$

$\Rightarrow \sqrt{2}x \pm \sqrt{3} = 0$

Length of the latus rectum of the hyperbola is $\frac{2 b^2}{a}$.

$\Rightarrow \frac{2 \times \left( \frac{5}{3} \right)}{\sqrt{\frac{5}{2}}} = \frac{10}{3}\sqrt{\frac{2}{5}}$

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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 27 Hyperbola
Exercise 27.1 | Q 3.5 | Page 13