Question

\\(\frac{x^2}{64} + \frac{y^2}{100} = 1\\)
the foci are located at
\\(\circ\\) (0, 8) and (0, -8)
\\(\circ\\) (0, 6) and (0, -6)
\\(\circ\\) (6, 0) and (-6, 0)

Explanation:

Step1: Identify the ellipse type

The standard form of an ellipse centered at the origin is \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}} = 1\) when \(a>b\), and the major axis is along the \(y\)-axis. For the given equation \(\frac{x^{2}}{64}+\frac{y^{2}}{100}=1\), we have \(a^{2} = 100\) (so \(a = 10\)) and \(b^{2}=64\) (so \(b = 8\)).

Step2: Calculate the distance from center to foci (\(c\))

The relationship for an ellipse is \(c^{2}=a^{2}-b^{2}\). Substituting the values, we get \(c^{2}=100 - 64=36\), so \(c = 6\).

Step3: Determine the foci coordinates

Since the major axis is along the \(y\)-axis, the foci are located at \((0,c)\) and \((0, - c)\). Substituting \(c = 6\), the foci are \((0,6)\) and \((0,-6)\).

Answer:

\(0, 6\) and \(0, -6\) (i.e., the option "O (0, 6) and (0, -6)")