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Find the eccentricity, centre , foci , and vertices of the following hyperbolas and draw their diagrams: $\large\frac{y^{2}}{9}-\frac{x^{2}}{25}=$$1$

1 Answer

  • 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}$$=1$
  • Foci $(0,\pm ae)$,centre $(0,0)$,vertices $(0,\pm a)$.
  • Transverse axis $y$-axis ($x=0$)
  • Conjugate axis $x$-axis ($y=0$)
  • End points of latus rectum $(\pm\large\frac{b^2}{a},$$ae),(\pm\large\frac{b^2}{a},$$-ae)$
  • Length of LR :$\large\frac{2b^2}{a}$
  • Directrices $y=\pm\large\frac{a}{e}$
Step 1:
$a=4,b=5$ and $x$-axis is the transverse axis.
Step 2:
Eccentricity $e=\sqrt{1+\large\frac{b^2}{a^2}}$
$\Rightarrow \sqrt{1+\large\frac{25}{9}}=\large\frac{\sqrt{34}}{3}$
Step 3:
Centre : Origin $O(0,0)$
Step 4:
Foci : $(0,\pm ae)$
$ae=3\times \large\frac{\sqrt{34}}{3}=$$\sqrt{34}$
$\Rightarrow (0,\pm\sqrt{34})$
Vertices : $(0,\pm a)$
$\Rightarrow (0,\pm 3)$
answered Jun 20, 2013 by sreemathi.v
edited Jun 20, 2013 by sreemathi.v

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