Answer
See graph
Work Step by Step
We are given the hyperbola:
$16x^2-y^2+64x-2y+67=0$
Bring the equation to the standard form:
$(16x^2+64x+64)-(y^2+2y+1)-64+1+67=0$
$16(x^2+4x+4)-(y+1)^2+4=0$
$16(x+2)^2-(y+1)^2+4=0$
$(y+1)^2-16(x+2)^2=4$
$\dfrac{(y+1)^2}{4}-\dfrac{16(x+2)^2}{4}=1$
$\dfrac{(y+1)^2}{4}-\dfrac{(x+2)^2}{\dfrac{1}{4}}=1$
The standard form of the equation is:
$\dfrac{(y-k)^2}{a^2}-\dfrac{(x-h)^2}{b^2}=1$
The transverse axis is parallel to the $y$-axis.
Determine $h,k,a,b,c$:
$h=-2$
$k=-1$
$a^2=4\Rightarrow a=\sqrt {4}=2$
$b^2=\dfrac{1}{4}\Rightarrow b=\sqrt {\dfrac{1}{4}}=\dfrac{1}{2}$
$c^2=a^2+b^2$
$c^2=4+\dfrac{1}{4}$
$c^2=\dfrac{17}{4}$
$c=\dfrac{\sqrt{17}}{2}$
Determine the centre:
$(h,k)=(-2,-1)$
Determine the coordinates of the vertices:
$(h,k-a)=(-2,-1-2)=(-2,-3)$
$(h,k+a)=(-2,-1+2)=(-2,1)$
Determine the coordinates of the foci:
$(h,k-c)=\left(-2,-1-\dfrac{\sqrt{17}}{2}\right)$
$(h,k+c)=\left(-2,-1+\dfrac{\sqrt{17}}{2}\right)$
Determine the asymptotes:
$y-k=\pm\dfrac{a}{b}(x-h)$
$y+1=\pm\dfrac{2}{\dfrac{1}{2}}(x+2)$
$y+1=\pm 4(x+2)$
Graph the hyperbola:
