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Find the equation of the hyperbola if foci: $(6 , -1 )$ and $(-4 , -1)$ and passing through the point $(4 , -1 )$

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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}$$=1$
  • Foci $(\pm ae,0)$,centre $(0,0)$,vertices $(\pm a,0)$.
  • Transverse axis $x$-axis ($y=0$)
  • Conjugate axis $y$-axis ($x=0$)
  • End points of latus rectum $(ae,\pm\large\frac{b^2}{a}),($$-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:
The centre is the midpoint of $FF'$.
(i.e) $C(\large\frac{6-4}{2},$$-1)=(1,-1)$
The transverse axis contains the foci.Its equation is $y=-1$ (parallel to $x$-axis).
The distance between the foci $FF'=2ae=10$
$\Rightarrow ae=5$---(1)
Step 2:
The point $P(4,-1)$ lies on the curve.
It lies on the transverse axis $y=-1$
It is a vertex of the hyperbola.
Therefore $CA=a=3$(length of semi transverse axis).
Step 3:
Substituting $a=3$ in (1),$e=\large\frac{5}{3}$
Therefore $b=\sqrt{\large\frac{25}{9}-\normalsize 1}$
$\Rightarrow b=3\times \large\frac{4}{3}$$=4$
Step 4:
The equation is of the form
Where $(h,k)$ is $C(1,-1)$
Therefore the equation is $\large\frac{(x-1)^2}{9}-\frac{(y+1)^2}{16}$$=1$
answered Jun 19, 2013 by sreemathi.v
edited Jun 19, 2013 by sreemathi.v

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