91影视

We have:

$9x^2-16y^2+18x+64y-199=0$

The equation is $9x^2-16y^2+18x+64y-199=0$.

Move the constant to the right:

$9x^2-16y^2+18x+64y=199$

Reorder the terms:

role="math" localid="1646037148259" $$\begin{gathered} 9x^2+18x-16y^2+64y=199 \\ 9\left(x^2+2x\right)-16y^2+64y=199 \end{gathered}$$

Add $9$to both sides:

role="math" localid="1646037239614" $$\begin{gathered} 9\left(x^2+2x+1\right)-16y^2+64y=199+9 \\ 9{\left(x+1\right)}^2-16\left(y^2-4y\right)=208 \end{gathered}$$

Add role="math" localid="1646037310735" $-64$to both sides:

role="math" localid="1646037360942" $$\begin{gathered} 9{\left(x+1\right)}^2-16\left(y^2-4y+4\right)=208-64 \\ 9{\left(x+1\right)}^2-16{\left(y-2\right)}^2=144 \\ \frac{9{\left(x+1\right)}^2}{144}-\frac{16{\left(y-2\right)}^2}{144}=\frac{144}{144} \\ \frac{{\left(x+1\right)}^2}{16}-\frac{{\left(y-2\right)}^2}{9}=1 \end{gathered}$$

Rewrite the equation:

$\frac{{\left(x-\left(-1\right)\right)}^2}{4^2}-\frac{{\left(y-2\right)}^2}{3^2}=1$

Observe that the equation is in the form of a horizontal hyperbola with the center located at $\left(-1,2\right)$.

The vertices of the hyperbola are:

$$\begin{gathered} \left(h-a,k\right),\left(h+a,k\right)=\left(-1-4,2\right),\left(-1+4,2\right) \\ =\left(-5,2\right),\left(3,2\right) \end{gathered}$$

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

role="math" localid="1646037695585" $$\begin{gathered} y=卤\frac{3}{4}\left(x-\left(-1\right)\right)+2 \\ y=\frac{3}{4}\left(x+1\right)+2,y=-\frac{3}{4}\left(x+1\right)+2 \end{gathered}$$

Connect the point with a line. The graph of the vertices $y=\frac{3}{4}\left(x+1\right)+2,y=-\frac{3}{4}\left(x+1\right)+2$.

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