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Find the equation of the hyperbola if centre : $(1 , 4 );$one of the foci $(6 , 4 )$ and the corresponding directrix is $x=\large\frac{9}{4}$

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  • 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}$$=1$
  • http://clay6.com/mpaimg/toolbox_10(i).png
  • Foci $(\pm ae,0)$,centre $(0,0)$,vertices $(\pm a,0)$.
  • Transverse axis $x$-axis ($y=0$)
  • Conjugate axis $y$-axis ($x=0$)
  • End points of latus rectum $(ae,\pm\large\frac{b^2}{a}),($$-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:
$C(1,4)$ and $F(6,4)$ be on $y=4$ which is the transverse axis (parallel to $x$-axis)
Step 2:
$\Rightarrow 6-\large\frac{9}{4}=\frac{15}{4}$----(2)
$\Rightarrow \large\frac{a}{e}=$$5-\large\frac{15}{4}=\frac{5}{4}$-----(3)
Step 3:
Eq(1) & Eq(2) gives $a^2=\large\frac{25}{4}$$\Rightarrow a=\large\frac{5}{2}$
$b=a\sqrt{e^2-1}=\large\frac{5}{2}$$\sqrt{4-1}$$=\large\frac{5\sqrt 3}{2}$
Step 4:
Since the transverse axis is parallel to the $x$-axis the equation is of the form
Where $(h,k)$ is the centre.
Therefore the equation is $\large\frac{(x-1)^2}{25/4}-\frac{(y-4)^2}{75/4}$$=1$
answered Jun 19, 2013 by sreemathi.v
edited Jun 19, 2013 by sreemathi.v

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