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

• 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$
$\Rightarrow a^2=2,b^2=8$
$\Rightarrow a=\sqrt 2,b=2\sqrt 2$
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,$2b=2\times 2\sqrt 2=4\sqrt 2$