Answer
See graph
Foci: $(-1,3-2\sqrt 5),(-1,3+2\sqrt 5)$
Work Step by Step
We are given the hyperbola:
$4x^2-y^2+8x+6y+11=0$
Bring the equation to the standard form:
$(4x^2+8x+4)-(y^2-6y+9)-4+9+11=0$
$4(x^2+2x+1)-(y^2-6y+9)+16=0$
$(y-3)^2-4(x+1)^2=16$
$\dfrac{(y-3)^2}{16}-\dfrac{4(x+1)^2}{16}=1$
$\dfrac{(y-3)^2}{16}-\dfrac{(x+1)^2}{4}=1$
The equation is in the form:
$\dfrac{(y-k)^2}{a^2}-\dfrac{(x-h)^2}{b^2}=1$
The transverse axis is parallel to the $y$-axis.
Determine $h,k,a,b,c$:
$h=-1$
$k=3$
$a^2=16\Rightarrow a=\sqrt {16}=4$
$b^2=4\Rightarrow b=\sqrt {4}=2$
$c^2=a^2+b^2$
$c^2=16+4$
$c^2=20$
$c=\sqrt{20}$
$c=2\sqrt 5$
The centre of the hyperbola is:
$(h,k)=(-1,3)$
Determine the coordinates of the vertices:
$(h,k-a)=(-1,3-4)=(-1,-1)$
$(h,k+a)=(-1,3+4)=(-1,7)$
Determine the coordinates of the foci:
$(h,k-c)=(-1,3-2\sqrt 5)$
$(h,k+c)=(-1,3+2\sqrt 5)$
Determine the asymptotes:
$y-k=\pm\dfrac{a}{b}(x-h)$
$y-3=\pm\dfrac{4}{2}(x+1)$
$y-3=\pm 2(x+1)$
Graph the hyperbola:
