Questions: What is the focus (are the foci) of the shape defined by the equation x^2/25 + y^2/9 = 1? (0,2) and (0,-2) (2,0) and (-2,0) (4,3) and (-4,-3) (4,0) and (-4,0) (0,4) and (0,-4)

Solution Steps

The given equation is

\[ \frac{x^{2}}{25} + \frac{y^{2}}{9} = 1 \]

This is the standard form of an ellipse, where \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). Here, \(a^2 = 25\) and \(b^2 = 9\).

Since \(a^2 > b^2\), the major axis is along the x-axis. Therefore, the ellipse is horizontally oriented.

For an ellipse, the distance from the center to each focus is given by \(c\), where \(c^2 = a^2 - b^2\).

Calculate \(c\):

\[ c^2 = a^2 - b^2 = 25 - 9 = 16 \]

\[ c = \sqrt{16} = 4 \]

The foci are located at \((\pm c, 0)\), which means the foci are at \((4, 0)\) and \((-4, 0)\).

Final Answer