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Find the equation of the hyperbola if centre: $(0 , 0 )$ length of the semi-transverse axis is $6; e=3$ and the transverse axis is parallel to $y$-axis.

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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{y^2}{a^2}-\frac{x^2}{b^2}$$=1$
  • http://clay6.com/mpaimg/2_toolbox10.jpg
  • Foci $(0,\pm ae)$,centre $(0,0)$,vertices $(0,\pm a)$.
  • Transverse axis $y$-axis ($x=0$)
  • Conjugate axis $x$-axis ($y=0$)
  • End points of latus rectum $(\pm\large\frac{b^2}{a},$$ae),(\pm\large\frac{b^2}{a},$$-ae)$
  • Length of LR :$\large\frac{2b^2}{a}$
  • Directrices $y=\pm\large\frac{a}{e}$
Step 1:
$C(0,0),a=6,e=3$,transverse axis is parallel to $y$-axis.
Equation is of the form $\large\frac{y^2}{a^2}-\frac{x^2}{b^2}$$=1$
$b^2=a^2(e^2=1)=36(9-1)=288$
Step 2:
The equation is $\large\frac{y^2}{36}-\frac{x^2}{288}$$=1$
answered Jun 19, 2013 by sreemathi.v
 

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