91Ó°ÊÓ

We have:

$9x^2-4y^2-18x+8y-31=0$

The equation of hyperbola is $9x^2-4y^2-18x+8y-31=0$.

Move the constant to the right:

$9x^2-4y^2-18x+8y=31$

Reorder the terms:

$9x^2-18x-4y^2+8y=31$

$9\left(x^2-2x\right)-4y^2+8y=31$

Add $9×1$to the both sides:

role="math" localid="1646050304266" $$\begin{gathered} 9\left(x^2-2x+1\right)-4y^2+8y=31+9×1 \\ 9{\left(x-1\right)}^2-4y^2+8y=31+9×1 \\ 9{\left(x-1\right)}^2-4y^2+8y=40 \end{gathered}$$

Add $-4×1$to both sides:

$$\begin{gathered} 9{\left(x-1\right)}^2-4\left(y^2+2y-1\right)=40-4×1 \\ 9{\left(x-1\right)}^2-4{\left(y-1\right)}^2=36 \\ \frac{9{\left(x-1\right)}^2}{36}-\frac{4{\left(y-1\right)}^2}{36}=\frac{36}{36} \\ \frac{{\left(x-1\right)}^2}{4}-\frac{{\left(y-1\right)}^2}{9}=1 \end{gathered}$$

Rewrite the equation:

$\frac{{\left(x-1\right)}^2}{2^2}-\frac{{\left(y-1\right)}^2}{3^2}=1$

Observe that the equation is in the form of a horizontal hyperbola with the center located at $\left(1,1\right)$.

Substitute $a=2$and $b=3$into the $a,b,c$relationship of hyperbola:

$$\begin{gathered} c^2=2^2+3^2 \\ c=Â\pm \sqrt{13} \end{gathered}$$

Since $c$represents length, it has to be positive so discard the negative solution.

The foci of horizontal hyperbola is given by:

$\left(h+c,k\right),\left(h-c,k\right)$

$=\left(1+\sqrt{3},1\right),\left(1-\sqrt{3},1\right)$

So, the vertices of a horizontal parabola:

$\left(1-2,1\right),\left(1+2,1\right)=\left(-1,1\right),\left(3,1\right)$

Substitute the value into the formula $y=Â\pm \frac{a}{b}\left(x-h\right)+k:$

role="math" localid="1646050847418" $y=Â\pm \frac{3}{2}\left(x-1\right)+1$

role="math" localid="1646050887370" $$\begin{gathered} y=\frac{3}{2}x-\frac{1}{2}, \\ y=-\frac{3}{2}x+\frac{5}{2} \end{gathered}$$

Connect the point with a line. The graph of the vertices $y=\frac{3}{2}x-\frac{1}{2},y=-\frac{3}{2}x+\frac{5}{2}$.

Draw the hyperbola passing through the vertices and considering the asymptotes. The graph of the function is shown below: