Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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Find the Eccentricity, Coordinates of the Foci, Equation of Directrice and Length of the Latus-rectum of the Hyperbola .9x2 − 16y2 = 144 - Mathematics

Answer in Brief

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

9x2 − 16y2 = 144

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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 = \[\frac{2 \times 9}{4} = \frac{9}{2}\]

  Is there an error in this question or solution?
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APPEARS IN

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