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Find the eccentricity, centre , foci , and vertices of the following hyperbolas and draw their diagrams: $25x^{2}-16y^{2}=400$

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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}$$=1$
  • http://clay6.com/mpaimg/6_toolbox_10(i).png
  • Foci $(\pm ae,0)$,centre $(0,0)$,vertices $(\pm a,0)$.
  • Transverse axis $x$-axis ($y=0$)
  • Conjugate axis $y$-axis ($x=0$)
  • End points of latus rectum $(ae,\pm\large\frac{b^2}{a}),($$-ae,\pm\large\frac{b^2}{a})$
  • Length of LR :$\large\frac{2b^2}{a}$
  • Directrices $y=\pm\large\frac{a}{e}$
Step 1:
$25x^2-16y^2=400$
The above equation is divided by $400$ we get
$\large\frac{x^2}{16}-\large\frac{y^2}{25}$$=1$
$a^2=16,b^2=25$
$\Rightarrow a=4,b=5$ and the axis is the transverse axis.
Step 2:
Ecentricity $e=\sqrt{1+\large\frac{b^2}{a^2}}=\sqrt{1+\large\frac{25}{16}}=\large\frac{\sqrt{41}}{4}$
Step 3:
Centre : Origin $O(0,0)$
Step 4:
Foci : $(\pm ae,0)$
$ae=4\times \large\frac{\sqrt{41}}{4}$
Foci $(\pm \sqrt{41},0)$
Step 5:
Vertices : $(\pm a,0)$
$\Rightarrow (\pm 4,0)$
answered Jun 25, 2013 by sreemathi.v
 

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