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Question
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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