Answer
$\frac{(x-3)^2}{9}+\frac{(y+\frac{1}{2})^2}{4}=1$, ellipse, center $C(3, -\frac{1}{2})$. foci $F(3\pm\sqrt 5, -\frac{1}{2})$, vertices $V_1(0, -\frac{1}{2})$ and $V_2(6, -\frac{1}{2})$.
See graph.
Work Step by Step
Step 1. Rewrite the equation as $16(x^2-6x+9)+36(y^2+y+\frac{1}{4})=16\times9+9-9$ or $16(x-3)^2+36(y+\frac{1}{2}^2)=144$
Step 2. Divide both sides by 144 to get $\frac{(x-3)^2}{9}+\frac{(y+\frac{1}{2})^2}{4}=1$ which represents an ellipse with $a=3, b=2$.
Step 3. The center can be found at $C(3, -\frac{1}{2})$.
Step 4. To find the foci, use the relation $c^2=a^2-b^2$ to get $c=\sqrt {9-4}=\sqrt 5$
Step 5. The original foci are $(\pm\sqrt 5, 0)$, and the shifted foci are $F(3\pm\sqrt 5, -\frac{1}{2})$
Step 6. The original vertices are $(\pm3, 0)$, and the shifted vertices are $V(3\pm3, -\frac{1}{2})$ or $V_1(0, -\frac{1}{2})$ and $V_2(6, -\frac{1}{2})$.
Step 7. See graph.
