Answer
$\frac{(x-1)^2}{81}+\frac{(y-2)^2}{56}=1$
Work Step by Step
Step 1. From the given conditions, we have
$a=\frac{10+8}{2}=9, c=\frac{6+4}{2}=5$, thus $b^2=9^2-5^2=56$
Step 2. The center is at $(\frac{10-8}{2}, 2)$or $(1,2)$; thus we can write the equation as $\frac{(x-1)^2}{81}+\frac{(y-2)^2}{56}=1$
