To determine the distance between the foci of the given ellipse, we first need to identify the parameters of the ellipse from its standard equation.
The given equation of the ellipse is:
[tex]\[
\frac{x^2}{16} + \frac{y^2}{9} = 1
\][/tex]
This equation is in the standard form:
[tex]\[
\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1
\][/tex]
where [tex]\( a^2 = 16 \)[/tex] and [tex]\( b^2 = 9 \)[/tex].
From this, we can find the semi-major and semi-minor axes:
[tex]\[
a = \sqrt{16} = 4
\][/tex]
[tex]\[
b = \sqrt{9} = 3
\][/tex]
Next, we need to determine the focal distance ([tex]\( c \)[/tex]) from the center of the ellipse to each focus. For an ellipse, the relationship between [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] is given by:
[tex]\[
c^2 = a^2 - b^2
\][/tex]
Substituting in the values we have:
[tex]\[
c^2 = 4^2 - 3^2 = 16 - 9 = 7
\][/tex]
[tex]\[
c = \sqrt{7} \approx 2.6457513110645907
\][/tex]
The distance between the foci is twice the focal distance ([tex]\( 2c \)[/tex]):
[tex]\[
2c = 2 \times \sqrt{7} \approx 2 \times 2.6457513110645907 = 5.291502622129181
\][/tex]
Therefore, the distance between the foci of the ellipse described by the equation [tex]\(\frac{x^2}{16}+\frac{y^2}{9}=1\)[/tex] is approximately [tex]\( 5.291502622129181 \)[/tex].
