The standard form that applies to the given equation is [latex]\dfrac{{y}^{2}}{{a}^{2}}-\dfrac{{x}^{2}}{{b}^{2}}=1[/latex]. Thus, the transverse axis is on the y-axis

The coordinates of the vertices are [latex]\left(0,\pm a\right)=\left(0,\pm \sqrt{64}\right)=\left(0,\pm 8\right)[/latex]

The coordinates of the co-vertices are [latex]\left(\pm b,0\right)=\left(\pm \sqrt{36},\text{ }0\right)=\left(\pm 6,0\right)[/latex]

The coordinates of the foci are [latex]\left(0,\pm c\right)[/latex], where [latex]c=\pm \sqrt{{a}^{2}+{b}^{2}}[/latex]. Solving for [latex]c[/latex] we have

[latex]\begin{align} c&=\pm \sqrt{{a}^{2}+{b}^{2}} \\ &=\pm \sqrt{64+36} \\ &=\pm \sqrt{100} \\ &=\pm 10 \end{align}[/latex]

Therefore, the coordinates of the foci are [latex]\left(0,\pm 10\right)[/latex]

The equations of the asymptotes are

[latex]\begin{align} y=\pm \frac{a}{b}x \\ y=\pm \frac{8}{6}x \\ y=\pm \frac{4}{3}x \end{align}[/latex]

Plot and label the vertices and co-vertices, and then sketch the central rectangle. Sides of the rectangle are parallel to the axes and pass through the vertices and co-vertices. Sketch and extend the diagonals of the central rectangle to show the asymptotes. The central rectangle and asymptotes provide the framework needed to sketch an accurate graph of the hyperbola. Label the foci and asymptotes, and draw a smooth curve to form the hyperbola.

Start by expressing the equation in standard form. Group terms that contain the same variable, and move the constant to the opposite side of the equation.

[latex]\left(9{x}^{2}-36x\right)-\left(4{y}^{2}+40y\right)=388[/latex]

Factor the leading coefficient of each expression.

[latex]9\left({x}^{2}-4x\right)-4\left({y}^{2}+10y\right)=388[/latex]

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

[latex]9\left({x}^{2}-4x+4\right)-4\left({y}^{2}+10y+25\right)=388+9\cdot4 - 4\cdot25[/latex]

Rewrite as perfect squares.

[latex]9{\left(x - 2\right)}^{2}-4{\left(y+5\right)}^{2}=324[/latex]

Divide both sides by the constant term to place the equation in standard form.

[latex]\dfrac{{\left(x - 2\right)}^{2}}{36}-\dfrac{{\left(y+5\right)}^{2}}{81}=1[/latex]

The standard form that applies to the given equation is [latex]\dfrac{{\left(x-h\right)}^{2}}{{a}^{2}}-\dfrac{{\left(y-k\right)}^{2}}{{b}^{2}}=1[/latex], where [latex]{a}^{2}=36[/latex] and [latex]{b}^{2}=81[/latex], or [latex]a=6[/latex] and [latex]b=9[/latex]. Thus, the transverse axis is parallel to the x-axis. It follows that:

the center of the ellipse is [latex]\left(h,k\right)=\left(2,-5\right)[/latex]

• the coordinates of the vertices are [latex]\left(h\pm a,k\right)=\left(2\pm 6,-5\right)[/latex], or [latex]\left(-4,-5\right)[/latex] and [latex]\left(8,-5\right)[/latex]
• the coordinates of the co-vertices are [latex]\left(h,k\pm b\right)=\left(2,-5\pm 9\right)[/latex], or [latex]\left(2,-14\right)[/latex] and [latex]\left(2,4\right)[/latex]
• the coordinates of the foci are [latex]\left(h\pm c,k\right)[/latex], where [latex]c=\pm \sqrt{{a}^{2}+{b}^{2}}[/latex]. Solving for [latex]c[/latex], we have

[latex]c=\pm \sqrt{36+81}=\pm \sqrt{117}=\pm 3\sqrt{13}[/latex]

Therefore, the coordinates of the foci are [latex]\left(2 - 3\sqrt{13},-5\right)[/latex] and [latex]\left(2+3\sqrt{13},-5\right)[/latex].

The equations of the asymptotes are [latex]y=\pm \frac{b}{a}\left(x-h\right)+k=\pm \frac{3}{2}\left(x - 2\right)-5[/latex].

Next, we plot and label the center, vertices, co-vertices, foci, and asymptotes and draw smooth curves to form the hyperbola.