The function given in the question is
[tex]9x^2+y^2+18x-12y-180=0[/tex]Rearrange the function above connecting similar terms
[tex]\begin{gathered} 9x^2+y^2+18x-12y-180=0 \\ 9x^2+18x+y^2-12y=180 \end{gathered}[/tex]Factorize the equation in brackets
[tex]\begin{gathered} 9x^2+18x+y^2-12y=180 \\ 9(x^2+2x)+1(y^2-12y)=180 \\ \end{gathered}[/tex]Solve the brackets using completing the square method but multiplying the coefficient of x and y by 1/2 and then square
[tex]\begin{gathered} 9(x^2+2x)+1(y^2-12y)=180 \\ 9(x^2+2x+(2\times\frac{1}{2})^2)+1(y^2-12y+(-12\times\frac{1}{2})^2=180+(2\times\frac{1}{2})^2+(-12\times\frac{1}{2})^2 \end{gathered}[/tex]By simplifying the equation above, we will have
[tex]\begin{gathered} 9(x^2+2x+(2\times\frac{1}{2})^2)+1(y^2-12y+(-12\times\frac{1}{2})^2=180+9(2\times\frac{1}{2})^2+1(-12\times\frac{1}{2})^2 \\ 9(x^2+2x+1)+1(y^2-12y+36)=180+9+36 \\ 9(x+1)^2+1(y-6)^2=225 \end{gathered}[/tex]Divide all through by 225
[tex]\begin{gathered} 9(x+1)^2+1(y-6)^2=225 \\ \frac{9(x+1)^2}{225}+\frac{1(y-6)^2}{225}=\frac{225}{225} \\ \frac{(x+1)^2}{25}+\frac{(y-6)^2}{225}=1 \end{gathered}[/tex]The general formula for the equation of an ellipse is
[tex]\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1[/tex]Where the center of the ellipse is
[tex](h,k)[/tex]By comparing coefficients, we will have that
[tex]\begin{gathered} x+1=x-h,y-6=y-k \\ x-x+h=-1,y-y+k=6 \\ h=-1,k=6 \end{gathered}[/tex]The centre of the ellipse is
[tex]\begin{gathered} (h,k)=(-1,6) \\ \text{hence,} \\ h+k=-1+6 \\ h+k=5 \end{gathered}[/tex]Therefore,
The value of h+k = 5
