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Find the equation and the length of the transverse and conjugate axis of the following hyperbola: $8y^{2}-2x^{2}=16$

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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}$$=1$
  • http://clay6.com/mpaimg/3_toolbox10.jpg
  • Foci $(0,\pm ae)$,centre $(0,0)$,vertices $(0,\pm a)$.
  • Transverse axis $y$-axis ($x=0$)
  • Conjugate axis $x$-axis ($y=0$)
  • End points of latus rectum $(\pm\large\frac{b^2}{a},$$ae),(\pm\large\frac{b^2}{a},$$-ae)$
  • Length of LR :$\large\frac{2b^2}{a}$
  • Directrices $y=\pm\large\frac{a}{e}$
Step 1:
$8y^2-2x^2=16$
The above equation is divided by $16$ we get
$\large\frac{y^2}{2}-\frac{x^2}{8}$$=1$
$\Rightarrow a^2=2,b^2=8$
$\Rightarrow a=\sqrt 2,b=2\sqrt 2$
Step 2:
Transverse axes : $y$-axes (i.e) $x=0$
Conjugate axes : $x$-axes (i.e) $y=0$
Length of transverse axis,$2a=2\sqrt 2$
Length of conjugate axis,$2b=2\times 2\sqrt 2=4\sqrt 2$
answered Jun 19, 2013 by sreemathi.v
 

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