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Find the equation and the length of the transverse and conjugate axis of the following hyperbola :$144x^{2}-25y^{2}=3600$

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Toolbox:
  • Standard forms of equation of the hyperbola with transverse axis $2a$,conjugate axis $2t$ with the negative sign associated with $b$ and $e=\sqrt{1+\large\frac{b^2}{a^2}},b=a\sqrt{e^2-1}$
  • $\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=1$
  • http://clay6.com/mpaimg/3_toolbox_10(i).png
  • Foci $(\pm ae,0)$,centre $(0,0)$,vertices $(\pm a,0)$.
  • Transverse axis $x$-axis ($y=0$)
  • Conjugate axis $y$-axis ($x=0$)
  • End points of latus rectum $(ae,\pm\large\frac{b^2}{a}),($$-ae,\pm\large\frac{b^2}{a})$
  • Length of LR :$\large\frac{2b^2}{a}$
  • Directrices $y=\pm\large\frac{a}{e}$
Step 1:
$144x^2-25y^2=3600$
The above equation is divided by $3600$
$\large\frac{x^2}{25}-\frac{y^2}{144}$$=1$
$\Rightarrow a^2=25,b^2=144$
$\Rightarrow a=5,b=12$
Step 2:
The transverse axis the $x$-axis,$y=0$
Conjugate axis is the $y$-axis,$x=0$
Length of transverse axis =$2a=2\times 5=10$
Length of conjugate axis =$2b=2\times 12=24$
answered Jun 19, 2013 by sreemathi.v
edited Jun 19, 2013 by sreemathi.v
 

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