Answer
$\frac{4(x-7)^2}{255}+\frac{(y-2)^2}{100}=1$
Work Step by Step
Step 1. Identify the given quantities: ellipse vertices $V_1(7,12)$ and $V_2(7,-8)$, passing point $P(1,8)$
Step 2. As the vertices are along a vertical line, we can assume a general equation as $\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1$
Step 3. The center is at the midpoint between the two vertices, we have $C(7, 2)$ thus $h=7, k=2$
Step 4. The distance between the two vertices is $2a=12+8=20$ thus $a=10$
Step 5. Plug-in the coordinates of the point $P(1,8)$, we have $\frac{(1-7)^2}{b^2}+\frac{(8-2)^2}{100}=1$
Step 6. Solve for $b^2$ from the equation above to get $b^2=\frac{225}{4}$
Step 7. Conclusion: the equation for the ellipse is $\frac{4(x-7)^2}{255}+\frac{(y-2)^2}{100}=1$
