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Find the equations of directrices ,latus rectums and length of latus rectum of the following hyperbolas: $4x^{2}-9y^{2}=576$

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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}$$=1$
  • http://clay6.com/mpaimg/5_toolbox_10(i).png
  • Foci $(\pm ae,0)$,centre $(0,0)$,vertices $(\pm a,0)$.
  • Transverse axis $x$-axis ($y=0$)
  • Conjugate axis $y$-axis ($x=0$)
  • End points of latus rectum $(ae,\pm\large\frac{b^2}{a}),($$-ae,\pm\large\frac{b^2}{a})$
  • End points of latus rectum $(ae,\pm\large\frac{b^2}{a}),($$-ae,\pm\large\frac{b^2}{a})$
  • Length of LR :$\large\frac{2b^2}{a}$
  • Directrices $y=\pm\large\frac{a}{e}$
Step 1:
$4x^2-9y^2=576$
The above equation is divided by $576$
$\large\frac{x^2}{144}-\frac{y^2}{64}$$=1$
$a^2=144,b^2=64$
$\Rightarrow a=12,b=8$
Step 2:
Eccentricity $e=\sqrt{1+\large\frac{b^2}{a^2}}=\sqrt{1+\large\frac{64}{144}}$
$\Rightarrow \sqrt{1+\large\frac{4}{9}}=\large\frac{\sqrt{13}}{3}$
Step 3:
Directrices : $x=\pm \large\frac{a}{e}$
$x=\pm \large\frac{12}{\sqrt{13}/3}$
$x=\pm \large\frac{36}{\sqrt{13}}$
Step 4:
Latus rectum : $x=\pm ae$
$x=\pm 12\times \large\frac{\sqrt{13}}{3}$
$x=\pm 4\sqrt {13}$
Step 5:
Length of $LR=\large\frac{2b^2}{a}$
$\Rightarrow \large\frac{2\times 64}{12}=\large\frac{32}{3}$
answered Jun 19, 2013 by sreemathi.v
edited Jun 19, 2013 by sreemathi.v
 

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