Answer
See graph
Work Step by Step
We are given the hyperbola:
$4x^2-9y^2+8x-18y-6=0$
Bring the equation to the standard form:
$(4x^2+8x+4)-(9y^2+18y+9)-4+9-6=0$
$4(x^2+2x+1)-9(y^2+2y+1)-1=0$
$4(x+1)^2-9(y+1)^2=1$
$\dfrac{(x+1)^2}{\dfrac{1}{4}}-\dfrac{(y+1)^2}{\dfrac{1}{9}}=1$
The equation is in the standard form:
$\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1$
The transverse axis is parallel to the $x$-axis.
Determine $h,k,a,b,c$:
$h=-1$
$k=-1$
$a^2=\dfrac{1}{4}\Rightarrow a=\sqrt {\dfrac{1}{4}}=\dfrac{1}{2}$
$b^2=\dfrac{1}{9}\Rightarrow b=\sqrt {\dfrac{1}{9}}=\dfrac{1}{3}$
$c^2=a^2+b^2$
$c^2=\dfrac{1}{4}+\dfrac{1}{9}$
$c^2=\dfrac{13}{36}$
$c=\sqrt{\dfrac{13}{36}}$
$c=\dfrac{\sqrt{13}}{6}$
The centre of the hyperbola is:
$(h,k)=(-1,-1)$
Determine the coordinates of the vertices:
$(h-a,k)=\left(-1-\dfrac{1}{2},-1\right)=\left(-\dfrac{3}{2},-1\right)$
$(h+a,k)=\left(-1+\dfrac{1}{2},-1\right)=\left(-\dfrac{1}{2},-1\right)$
Determine the coordinates of the foci:
$(h-c,k)=\left(-1-\dfrac{\sqrt{13}}{6},-1\right)$
$(h+c,k)=\left(-1+\dfrac{\sqrt{13}}{6},-1\right)$
Determine the asymptotes:
$y-k=\pm\dfrac{b}{a}(x-h)$
$y+1=\pm\dfrac{\dfrac{1}{3}}{\dfrac{1}{2}}(x+1)$
$y+1=\pm\dfrac{2}{3}(x+1)$
Graph the hyperbola:
