Given the equation,
[tex]-9x^2+72x+16y^2+16y+4=0[/tex]Complete squares as shown below,
[tex]\begin{gathered} -9x^2+72x-a^2=-(9x^2-72x+a^2)=-9(x^2-8x+b^2) \\ \end{gathered}[/tex]Thus,
[tex]\begin{gathered} \Rightarrow-9x^2+72x-a^2=-9(x^{}-4)^2 \\ \Rightarrow a^2=16\cdot9=144\Rightarrow a=12 \\ \Rightarrow-9x^2+72x-144=-9(x^{}-4)^2 \end{gathered}[/tex]Similarly,
[tex]\begin{gathered} 16y^2+16y=16(y^2+y) \\ \Rightarrow16(y+\frac{1}{2})^2=16(y^2+y+\frac{1}{4}) \end{gathered}[/tex]Therefore,
[tex]\begin{gathered} -9x^2+72x+16y^2+16y+4=0 \\ \Rightarrow-9(x-4)^2+16(y+\frac{1}{2})^2+4=-144+4 \\ \Rightarrow-9(x-4)^2+16(y+\frac{1}{2})^2=-144 \end{gathered}[/tex]Finally, the standard form is.
[tex]\begin{gathered} \Rightarrow-\frac{(x-4)^2}{16}+\frac{(y+\frac{1}{2})^2}{9}=-1 \\ \Rightarrow\frac{(x-4)^2}{16}-\frac{(y+\frac{1}{2})^2}{9}=1 \end{gathered}[/tex]As for the vertices, foci, and asymptotes,
[tex]\begin{gathered} c=\pm\sqrt[]{16+9}=\pm5 \\ \text{center:}(4,-\frac{1}{2}) \\ \Rightarrow\text{foci:}(4-5,-\frac{1}{2})_{},(4+5,-\frac{1}{2})_{} \\ \Rightarrow\text{foci:}(-1,-\frac{1}{2}),(9,-\frac{1}{2}) \end{gathered}[/tex]Foci: (-1,-1/2), (9,-1/2)
Vertices
[tex]\begin{gathered} \text{center:}(4,-\frac{1}{2}),\text{vertices:}(4\pm a,-\frac{1}{2}) \\ \text{vertices:}(4+4,-\frac{1}{2}),(4-4,-\frac{1}{2}) \\ \text{vertices:}(8,-\frac{1}{2}),(0,-\frac{1}{2}) \end{gathered}[/tex]Vertices: (8,-1/2), (0,-1/2)
Asymptotes:
[tex]\begin{gathered} y=\pm\frac{3}{4}(x-4)-\frac{1}{2} \\ \Rightarrow y=\frac{3}{4}x-\frac{7}{2} \\ \text{and} \\ y=-\frac{3}{4}x+\frac{5}{2} \end{gathered}[/tex]Asymptotes: y=3x/4-7/2 and y=-3x/4+5/2
