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Find the equation of the hyperbola if foci:$(0,\pm 8 )$; length of transverse axis is $12$ .

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1 Answer

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  • General form of standard hyperbola with centre $C(h,k)$,transverse axis $2a$,conjugate axis $2b$,$(b^2-ve)$ and with axes parallel to the coordinate axes.
  • (i) $\large\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}$$=1$
  • Transverse axis $y-k=0$,conjugate axis $x-h=0$.
  • (ii) $\large\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}$$=1$
  • Transverse axis $x-h=0$,conjugate axis $y-k=0$.
Step 1:
Foci lie on $y$-axis with origin as mid point.Therefore the transverse axis is the $y$-axis and the origin is the centre.
Step 2:
Distance between foci=16=$2ae$
$\Rightarrow ae=8$
Length of transverse axis =12=$2a$
$\Rightarrow a=6$
$e=\large\frac{8}{6}=\frac{4}{3}$
$b=6\sqrt{\large\frac{16}{9}-\normalsize 1}$
$\Rightarrow \large\frac{6\sqrt 7}{3}=$$2\sqrt 7$
Step 3:
The equation is of the form
$\large\frac{y^2}{a^2}-\frac{x^2}{b^2}$$=1$
$\large\frac{y^2}{36}-\frac{x^2}{28}$$=1$
answered Jun 19, 2013 by sreemathi.v
 

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