Answer
see graph
Work Step by Step
We are given the hyperbola:
$4x^2-y^2+32x+6y+39=0$
Bring the equation to the standard form:
$(4x^2+32x+64)-(y^2-6y+9)-64+9+39=0$
$4(x^2+8x+16)-(y-3)^2-16=0$
$4(x+4)^2-(y-3)^2=16$
$\dfrac{4(x+4)^2}{16}-\dfrac{(y-3)^2}{16}=1$
$\dfrac{(x+4)^2}{4}-\dfrac{(y-3)^2}{16}=1$
The standard form of the equation is:
$\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1$
The transverse axis is parallel to the $x$-axis.
Determine $h,k,a,b,c$:
$h=-4$
$k=3$
$a^2=4\Rightarrow a=\sqrt 4=2$
$b^2=16\Rightarrow b=\sqrt {16}=4$
$c^2=a^2+b^2$
$c^2=4+16$
$c^2=20$
$c=\sqrt{20}=2\sqrt 5$
The centre of the hyperbola is:
$(h,k)=(-4,3)$
Determine the coordinates of the vertices:
$(h-a,k)=(-4-2,3)=(-6,3)$
$(h+a,k)=(-4+2,3)=(-2,3)$
Determine the coordinates of the foci:
$(h-c,k)=(-4-2\sqrt{5},3)$
$(h+c,k)=(-4+2\sqrt{5},3)$
Determine the asymptotes:
$y-k=\pm\dfrac{b}{a}(x-h)$
$y-3=\pm\dfrac{4}{2}(x+4)$
$y-3=\pm 2(x+4)$
Graph the hyperbola:
