Question

what are the foci of the ellipse \\( \frac { x ^ { 2 } } { 65 } + \frac { y ^ { 2 } } { 40 } = 1 \\)?
write your answer in simplified, rationalized form.
\\( ( \square, \square ) \\) and \\( ( \square, \square ) \\)

Explanation:

Step1: Identify the form of the ellipse

The standard form of an ellipse is \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = 1\) (\(a>b\)). For the ellipse \(\frac{x^{2}}{65}+\frac{y^{2}}{40}=1\), we have \(a^{2}=65\), \(b^{2} = 40\).

Step2: Calculate \(c\) using the formula \(c^{2}=a^{2}-b^{2}\)

Substitute \(a^{2}=65\) and \(b^{2}=40\) into \(c^{2}=a^{2}-b^{2}\). Then \(c^{2}=65 - 40=25\), so \(c = 5\).

Step3: Determine the foci

Since the major - axis is along the \(x\) - axis (because \(a^{2}=x^{2}\) term), the foci of the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) are \((\pm c,0)\).

Answer:

\((5,0)\) and \((- 5,0)\)