Question

what are the foci of the ellipse \\( \frac { ( x - 1 ) ^ { 2 } } { 64 } + \frac { y ^ { 2 } } { 4 } = 1 \\)?
write your answer in simplified, rationalized form.
(, ) and (, )

Explanation:

Step1: Identify the form of the ellipse equation

The standard form of an ellipse is \(\frac{(x - h)^{2}}{a^{2}}+\frac{(y - k)^{2}}{b^{2}} = 1\) (\(a>b\)). Here, \(h = 1,k = 0,a^{2}=64\), so \(a = 8\), \(b^{2}=4\), so \(b = 2\).

Step2: Calculate \(c\)

Use the formula \(c^{2}=a^{2}-b^{2}\). Substitute \(a = 8\) and \(b = 2\) into it: \(c^{2}=64 - 4=60\), then \(c=\sqrt{60}=2\sqrt{15}\).

Step3: Find the foci

For an ellipse \(\frac{(x - h)^{2}}{a^{2}}+\frac{(y - k)^{2}}{b^{2}} = 1\) (\(a>b\)), the foci are \((h\pm c,k)\). Substitute \(h = 1,k = 0,c = 2\sqrt{15}\) into it. The foci are \((1 + 2\sqrt{15},0)\) and \((1-2\sqrt{15},0)\).

Answer:

\((1 + 2\sqrt{15},0)\) and \((1-2\sqrt{15},0)\)