Answer
$\frac{(x-4)^2}{4}-\frac{3(y)^2}{16}=1$
Work Step by Step
Step 1. Based on the shape and orientation of the curve, we can assume the equation as $\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$
Step 2. Identify the center as $(4,0)$, we have $h=4, k=0$
Step 3. Identify the vertices as $(2,0)$ and $(6,0)$ which gives $2a=6-2=4$ and $a=2$
Step 4. As the curve pass point $(0,4)$, we have $\frac{(0-4)^2}{2^2}-\frac{(4)^2}{b^2}=1$. Solve for $b^2$ to get $b^2=\frac{16}{3}$
Step 5. Conclusion: the equation of the curve is $\frac{(x-4)^2}{4}-\frac{3(y)^2}{16}=1$
