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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}$$=1 • http://clay6.com/mpaimg/1_5_toolbox10.jpg • Foci (0,\pm ae),centre (0,0),vertices (0,\pm a). • Transverse axis y-axis (x=0) • Conjugate axis x-axis (y=0) • End points of latus rectum (\pm\large\frac{b^2}{a},$$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}$$=1 • Transverse axis x-h=0,Conjugate axis y-k=0 Step 1: x^2-3y^2+6x+6y+18=0 x^2-3y^2+6x+6y=-18 Completing the squares, (x^2+6x+9)-3(y^2-2y+1)=-18+9-3 (x+3)^2-3(y-1)^2=-12 The above equation is divided by -12 we get \large\frac{(y-1)^2}{4}-\large\frac{(x+3)^2}{12}$$=1$
$a^2=4,b^2=12$
$\Rightarrow a=2,b=2\sqrt 3$
Step 2:
Shifting the origin to $(-1,3)$ by translation of axes.
$X=x+3$,$Y=y-1$
The equation reduces to the form $\large\frac{Y^2}{4}-\large\frac{X^2}{12}$$=1 The Y-axis is the transverse axis. Step 3: XY-axes : e=\sqrt{1+\large\frac{b^2}{a^2}} \Rightarrow \sqrt{1+\large\frac{12}{4}}=$$\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