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)$(-5,2),(-1,2)$