Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# Find the Eccentricity, Coordinates of the Foci, Equation of Directrice and Length of the Latus-rectum of the Hyperbola .9x2 − 16y2 = 144 - Mathematics

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

9x2 − 16y2 = 144

#### Solution

Equation of the hyperbola: $9 x^2 - 16 y^2 = 144$ This can be rewritten in the following way:

$\frac{x^2}{16} - \frac{y^2}{9} = 1$

This is the standard equation of a hyperbola, where  $a^2 = 16 \text { and }b^2 = 9$ .

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

$\Rightarrow 9 = 16( e^2 - 1)$

$\Rightarrow e^2 - 1 = \frac{9}{16}$

$\Rightarrow e^2 = \frac{25}{16}$

$\Rightarrow e = \frac{5}{4}$

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

$\left( \pm 5, 0 \right)$ .

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

$\Rightarrow x = \pm \frac{4}{\frac{5}{4}}$

$\Rightarrow 5x \pm 16 = 0$

Length of the latus rectum of the hyperbola is  $\frac{2 b^2}{a}$ Length of the latus rectum = $\frac{2 \times 9}{4} = \frac{9}{2}$

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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 27 Hyperbola
Exercise 27.1 | Q 3.1 | Page 13