Calculus: Early Transcendentals 8th Edition

Given: $16x^{2}+9y^{2}-36y=108$ By completing the square, we get $16x^{2}+9(y^{2}-4y+4)=108+9(4)$ $16x^{2}+9(y-2)^{2}=144$ Divide by $144$. $\frac{x^{2}}{9}+\frac{(y-2)^{2}}{16}=1$ which is the standard form of the equation of an Ellipse. $\frac{(x-h)^{2}}{b^{2}}+\frac{(y-k)^{2}}{a^{2}}=1$ Here, $(h,k)=(0,2)$