To find the foci of the ellipse given by the equation [tex]\(\frac{x^2}{144}+\frac{y^2}{100}=1\)[/tex], we need to follow these steps:
1. Identify the values of [tex]\(a^2\)[/tex] and [tex]\(b^2\)[/tex] from the standard form of the ellipse equation [tex]\(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)[/tex].
2. Compare the given equation [tex]\(\frac{x^2}{144} + \frac{y^2}{100} = 1\)[/tex] with the standard form to identify [tex]\(a^2\)[/tex] and [tex]\(b^2\)[/tex]:
- [tex]\(a^2 = 144\)[/tex]
- [tex]\(b^2 = 100\)[/tex]
3. Calculate [tex]\(a\)[/tex] and [tex]\(b\)[/tex]:
- [tex]\(a = \sqrt{144} = 12\)[/tex]
- [tex]\(b = \sqrt{100} = 10\)[/tex]
4. Determine the relationship between [tex]\(a\)[/tex] and [tex]\(b\)[/tex]:
- Since [tex]\(a > b\)[/tex], the major axis is along the x-axis, and the foci are at points [tex]\((\pm c, 0)\)[/tex].
5. Calculate [tex]\(c\)[/tex] using the formula [tex]\(c = \sqrt{a^2 - b^2}\)[/tex]:
- [tex]\(c = \sqrt{144 - 100} = \sqrt{44} \approx 6.6332495807108\)[/tex]
6. Hence, the foci of the ellipse are at:
- [tex]\((\pm c, 0) = (\pm 6.6332495807108, 0)\)[/tex]
Therefore, the foci of the ellipse are located at [tex]\((\pm 6.6332495807108, 0)\)[/tex]. Comparing this with the given options, the correct one is [tex]\(( \pm 2 \sqrt{11}, 0)\)[/tex], since [tex]\(2 \sqrt{11} = \sqrt{44} \approx 6.6332495807108\)[/tex].
Thus, the correct answer is:
[tex]\[
(\pm 2 \sqrt{11}, 0)
\][/tex]
