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Find the eccentricity, centre , foci , and vertices of the following hyperbolas and draw their diagrams: $x^{2}-4y^{2}+6x+16y-11=0$

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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{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 :$\large\frac{2b^2}{a}$
  • 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 squares,
The above equation is divided by 4 we get
Step 2:
Shifting the origin to $(-3,2)$ by translation of axes.
The equation becomes $\large\frac{X^2}{4}-\frac{Y^2}{1}$$=1$
The transverse axis is the $x$-axis.
Step 3:
$XY$-axes :
Eccentricity $e=\sqrt{1+\large\frac{b^2}{a^2}}$
$\Rightarrow \sqrt{1+\large\frac{1}{4}}=\large\frac{\sqrt 5}{2}$
Centre :$(0,0)$
Foci : $(\pm ae,0)$
$ae=2\large\frac{\sqrt 5}{2}$
$\Rightarrow \sqrt 5$
$\Rightarrow (\pm \sqrt 5,0)$
Vertices $(\pm a,0)$
$(\pm 2,0)$
$xy$-axis :
Centre :$(-3,2)$
Foci : $(-3\pm \sqrt 5,2)$
Vertices : $(-3\pm 2,2)$
answered Jun 20, 2013 by sreemathi.v
edited Jun 20, 2013 by sreemathi.v

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