## Calculus (3rd Edition)

The answers: $\begin{array}{*{20}{c}} {}&{standard}&{translated}\\ {vertices\left( {focal} \right)}&{\left( { \pm a,0} \right) = \left( { \pm 5,0} \right)}&{\left( {7,4} \right),\left( { - 3,4} \right)}\\ {vertices\left( {conjugate} \right)}&{\left( {0, \pm b} \right) = \left( {0, \pm 4} \right)}&{\left( {2,8} \right),\left( {2,0} \right)}\\ {foci}&{\left( { \pm c,0} \right) = \left( { \pm 3,0} \right)}&{\left( {5,4} \right),\left( { - 1,4} \right)}\\ {center}&{\left( {0,0} \right)}&{\left( {2,4} \right)} \end{array}$
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: $\begin{array}{*{20}{c}} {}&{standard}&{translated}\\ {vertices\left( {focal} \right)}&{\left( { \pm a,0} \right) = \left( { \pm 5,0} \right)}&{\left( {7,4} \right),\left( { - 3,4} \right)}\\ {vertices\left( {conjugate} \right)}&{\left( {0, \pm b} \right) = \left( {0, \pm 4} \right)}&{\left( {2,8} \right),\left( {2,0} \right)}\\ {foci}&{\left( { \pm c,0} \right) = \left( { \pm 3,0} \right)}&{\left( {5,4} \right),\left( { - 1,4} \right)}\\ {center}&{\left( {0,0} \right)}&{\left( {2,4} \right)} \end{array}$