Answer
The answers:
$\begin{array}{*{20}{c}}
{}&{standard}&{translated}\\
{vertices\left( {focal} \right)}&{\left( { \pm a,0} \right) = \left( { \pm 5,0} \right)}&{\left( {7,4} \right),\left( { - 3,4} \right)}\\
{vertices\left( {conjugate} \right)}&{\left( {0, \pm b} \right) = \left( {0, \pm 4} \right)}&{\left( {2,8} \right),\left( {2,0} \right)}\\
{foci}&{\left( { \pm c,0} \right) = \left( { \pm 3,0} \right)}&{\left( {5,4} \right),\left( { - 1,4} \right)}\\
{center}&{\left( {0,0} \right)}&{\left( {2,4} \right)}
\end{array}$
Work Step by Step
Write
$16{x^2} + 25{y^2} - 64x - 200y + 64 = 0$
$16\left( {{x^2} - 4x} \right) + 25\left( {{y^2} - 8y} \right) + 64 = 0$
$16\left( {{x^2} - 4x + 4} \right) - 64 + 25\left( {{y^2} - 8y + 16} \right) - 400 + 64 = 0$
$16{\left( {x - 2} \right)^2} + 25{\left( {y - 4} \right)^2} = 400$
Divide both sides by $400$ gives
$\frac{1}{{25}}{\left( {x - 2} \right)^2} + \frac{1}{{16}}{\left( {y - 4} \right)^2} = 1$
${\left( {\frac{{x - 2}}{5}} \right)^2} + {\left( {\frac{{y - 4}}{4}} \right)^2} = 1$
From this equation we see that this is an ellipse with $a=5$, $b=4$ and centered at $C=\left(2,4\right)$.
By Theorem 1, we get
$c = \sqrt {{a^2} - {b^2}} = \sqrt {25 - 16} = 3$
Translated vertices:
Focal vertices:
$\left( {5,0} \right) + \left( {2,4} \right) = \left( {7,4} \right)$
$\left( { - 5,0} \right) + \left( {2,4} \right) = \left( { - 3,4} \right)$
Conjugate vertices:
$\left( {0,4} \right) + \left( {2,4} \right) = \left( {2,8} \right)$
$\left( {0, - 4} \right) + \left( {2,4} \right) = \left( {2,0} \right)$
Translated foci:
$\left( {3,0} \right) + \left( {2,4} \right) = \left( {5,4} \right)$
$\left( { - 3,0} \right) + \left( {2,4} \right) = \left( { - 1,4} \right)$
We summarize the results in the following table:
$\begin{array}{*{20}{c}}
{}&{standard}&{translated}\\
{vertices\left( {focal} \right)}&{\left( { \pm a,0} \right) = \left( { \pm 5,0} \right)}&{\left( {7,4} \right),\left( { - 3,4} \right)}\\
{vertices\left( {conjugate} \right)}&{\left( {0, \pm b} \right) = \left( {0, \pm 4} \right)}&{\left( {2,8} \right),\left( {2,0} \right)}\\
{foci}&{\left( { \pm c,0} \right) = \left( { \pm 3,0} \right)}&{\left( {5,4} \right),\left( { - 1,4} \right)}\\
{center}&{\left( {0,0} \right)}&{\left( {2,4} \right)}
\end{array}$
