To find the foci of the ellipse given by the equation 25x² + 16y² = 400, we first need to rewrite the equation in the standard form of an ellipse.
The standard form of an ellipse is given by:
\( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \), where:
- \( (h, k) \) is the center of the ellipse.
- \( a \) is the distance from the center to the vertices along the major axis.
- \( b \) is the distance from the center to the vertices along the minor axis.
First, we divide the entire equation by 400 to normalize it:
\( \frac{25x^2}{400} + \frac{16y^2}{400} = 1 \)
This simplifies to:
\( \frac{x^2}{16} + \frac{y^2}{25} = 1 \)
From this equation, we can see that:
- \( a^2 = 25 \) (thus, \( a = 5 \)), and
- \( b^2 = 16 \) (thus, \( b = 4 \)).
Since \( a^2 > b^2 \), the major axis is along the y-axis.
The center of the ellipse is at the origin, \( (0, 0) \).
To find the foci, we use the formula:
\( c = \sqrt{a^2 – b^2} \)
Calculating \( c \):
\( c = \sqrt{25 – 16} = \sqrt{9} = 3 \)
The foci of the ellipse are located at \( (0, c) \) and \( (0, -c) \), which gives us:
Foci: (0, 3) and (0, -3)