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Find the eccentricity, centre , foci , and vertices of the following hyperbolas and draw their diagrams: $x^{2}-3y^{2}+6x+6y+18=0$

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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{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},$$-ae)$
  • End points of latus rectum $(\pm\large\frac{b^2}{a},$$ae),(\pm\large\frac{b^2}{a},$$-ae)$
  • Length of LR :$\large\frac{2b^2}{a}$
  • Directrices $y=\pm\large\frac{a}{e}$
  • General form of standard hyperbolic with centre $C(h,k)$,transverse axis $2a$,conjugate axis $2b$,$(b^2$ -ve) and with axes parallel to the coordinate axes.
  • (i) $\large\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}$$=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}$$=1$
  • Transverse axis $x-h=0$,Conjugate axis $y-k=0$
Step 1:
Completing the squares,
The above equation is divided by $-12$ we get
$\Rightarrow a=2,b=2\sqrt 3$
Step 2:
Shifting the origin to $(-1,3)$ by translation of axes.
The equation reduces to the form $\large\frac{Y^2}{4}-\large\frac{X^2}{12}$$=1$
The $Y$-axis is the transverse axis.
Step 3:
$XY$-axes :
$\Rightarrow \sqrt{1+\large\frac{12}{4}}=$$\sqrt4=2$
Centre :$(0,0)$
Foci : $(0,\pm ae)$
$\Rightarrow (0,\pm 2\times 2)$
$\Rightarrow (0,\pm 4)$
Vertices : $(0,\pm a)$
$\Rightarrow (0,\pm 2)$
Step 4:
Centre : $(1,-3)$
Foci : $(1,-3\pm 4)$
(i.e) $(1,7),(1,1)$
Vertices : $(1,-3\pm 2)$
(i.e) $(1,-5),(1,-1)$
answered Jun 25, 2013 by sreemathi.v

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