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Find the equation of the hyperbola if foci: $(\pm 3 , 5 ); e=3$

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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}$$=1$
  • http://clay6.com/mpaimg/2_toolbox_10(i).png
  • Foci $(\pm ae,0)$,centre $(0,0)$,vertices $(\pm a,0)$.
  • Transverse axis $x$-axis ($y=0$)
  • Conjugate axis $y$-axis ($x=0$)
  • End points of latus rectum $(ae,\pm\large\frac{b^2}{a}),($$-ae,\pm\large\frac{b^2}{a})$
  • Length of LR :$\large\frac{2b^2}{a}$
  • Directrices $y=\pm\large\frac{a}{e}$
  • 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:
Distance between the foci =$2ae=6$
$\Rightarrow ae=3$
Now $e=3$
Therefore $3a=3$
$a=1$
$b=a\sqrt{e^2-1}$
$\Rightarrow \sqrt{9-1}=2\sqrt 2$
Step 2:
The transverse axis is $y=5$(parallel to $x$-axis)
Since it contains the foci.
The centre is the mid point of $FF'$(i.e) $C(0,5)$
Step 3:
The equation is of the form
$\large\frac{(x-h)^2}{a^2}-\large\frac{(y-k)^2}{b^2}$$=1$
Where $(h,k)$ is $(0,5)$
$\large\frac{x^2}{1}-\large\frac{(y-5)^2}{8}$$=1$
answered Jun 19, 2013 by sreemathi.v
edited Jun 19, 2013 by sreemathi.v
 

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