Answer
$\frac{(x-2)^2}{4}-\frac{(y-4)^2}{5}=1$
Work Step by Step
Step 1. Identify the given quantities: hyperbola center $C(2,4)$, foci $F_1(2,1)$ and $F_2(2,7)$, vertices $V_1(2,6)$ and $V_2(2,2)$.
Step 2. As both the foci and vertices are horizontal, we can assume the hyperbola as $\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$
Step 3. With the center at $(2,4)$, we have $h=2, k=4$
Step 4. The distance between the foci is $2c=7-1=6$, thus $c=3$
Step 5. The distance between the vertices is $2a=6-2=4$, thus $a=2$
Step 6. Use the relation $b^2=c^2-a^2=9-4=5$, we have $b=\sqrt 5$
Step 7. Conclusion: the equation for the hyperbola is $\frac{(x-2)^2}{4}-\frac{(y-4)^2}{5}=1$
