Question

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

16x² − 9y² = −144

Answer 1

 Equation of the hyperbola: \[16 x^2 - 9 y^2 = - 144\]

This can be rewritten in the following way:

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

This is the standard equation of a hyperbola, where 

\[a^2 = 9 \text { and } b^2 = 16\].

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

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

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

\[ \Rightarrow e^2 = \frac{25}{16}\]

\[ \Rightarrow e = \frac{5}{4}\]

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

\[\left( 0, \pm 5 \right)\] .

Equation of the directrices: \[y = \pm \frac{a}{e}\]

\[\Rightarrow y = \pm \frac{4}{\frac{5}{4}}\]

\[ \Rightarrow 5y \pm 16 = 0\]

Length of the latus rectum of the hyperbola = \[\frac{2 a^2}{b}\]

 

Length of the latus rectum = \[\frac{2 \times 9}{4} = \frac{9}{2}\].