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# Find the equation of the hyperbola if centre: $(0 , 0 )$ length of the semi-transverse axis is $5; e=\large\frac{7}{5}$ and the conjugate axis is along x-axis.

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}$$=1 • http://clay6.com/mpaimg/toolbox10.jpg • Foci (0,\pm ae),centre (0,0),vertices (0,\pm a). • Transverse axis y-axis (x=0) • Conjugate axis x-axis (y=0) • End points of latus rectum (\pm\large\frac{b^2}{a},$$ae),(\pm\large\frac{b^2}{a},$$-ae) • Length of LR :\large\frac{2b^2}{a} • Directrices y=\pm\large\frac{a}{e} Step 1: a=5,e=\large\frac{7}{5} Equation is of the form \large\frac{y^2}{a^2}-\frac{x^2}{b^2}$$=1$
$b^2=a^2(e^2-1)$
$\Rightarrow 25\big(\large\frac{49}{25}$$-1\bigg)$$=24$
$b^2=24$
Step 2:
Therefore the equation is $\large\frac{y^2}{25}-\frac{x^2}{24}$$=1$
edited Jun 19, 2013