Answer
$\frac{(x-3)^{2}}{29} + \frac{(y+4)^{2}}{25} = 1$
Work Step by Step
General Equation of an ellipse is: $\frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1$
Foci at (1, -4) and (5,-4); midpoint of the foci is center, which is at (3, -4). h = 3 and k = -4
c (distance of focus to center) is 2
Thus we know this: $\frac{(x-3)^{2}}{a^{2}} + \frac{(y+4)^{2}}{b^{2}} = 1$
Substitute (3,1) as given in problem
$\frac{(3-3)^{2}}{a^{2}} + \frac{(1+4)^{2}}{b^{2}} = 1$
$\frac{25}{b^{2}} = 1$
Thus $b^{2} = 25$
$c^{2} = a^{2} - b^{2}$
$ 4 = a^{2} - 25$
Thus $a^{2} = 29$
Final equation is $\frac{(x-3)^{2}}{29} + \frac{(y+4)^{2}}{25} = 1$
