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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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A)
Toolbox:
• 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}$$=1 • http://clay6.com/mpaimg/8_toolbox_10(i).png • Foci (\pm ae,0),centre (0,0),vertices (\pm a,0). • Transverse axis x-axis (y=0) • Conjugate axis y-axis (x=0) • End points of latus rectum (ae,\pm\large\frac{b^2}{a}),($$-ae,\pm\large\frac{b^2}{a})$
• Length of LR :$\large\frac{2b^2}{a}$
• 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:
$x^2-4y^2+6x+16y-11=0$
$x^2+6x-4y^2+16y=11$
Completing squares,
$(x^2+6x+9)-4(y^2-4y+4)=11+9-16$
$(x+3)^2-4(y-2)^2=4$
The above equation is divided by 4 we get
$\large\frac{(x+3)^2}{4}-\large\frac{(y-2)^2}{1}=$$1 a^2=4,b^2=1 a=2,b=1 Step 2: Shifting the origin to (-3,2) by translation of axes. X=x+3,Y=y-2 x=X-3,y=Y+2 The equation becomes \large\frac{X^2}{4}-\frac{Y^2}{1}$$=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 : $(-3\pm 2,2)$
(i.e)$(-5,2),(-1,2)$