23. Find the foci of the ellipse. 3D1 169 144

Algebra and Trigonometry (6th Edition)
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ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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**Question 23: Find the foci of the ellipse.**

\[\frac{x^2}{169} + \frac{y^2}{144} = 1\]

Options:
- (a) (6, 0) and (-6, 0)
- (b) (5, 0) and (-5, 0)
- (c) (7, 0) and (-7, 0)
- (d) (8, 0) and (-8, 0)

**Explanation:**
This problem asks to determine the foci of an ellipse given in its standard form equation. The equation of the ellipse is \(\frac{x^2}{169} + \frac{y^2}{144} = 1\), indicating it is a horizontal ellipse because 169 (the larger denominator) is associated with \(x^2\).

To find the foci, we use the formula for c, where \(c = \sqrt{a^2 - b^2}\). Here, \(a^2 = 169\) and \(b^2 = 144\). Calculating c:

\[c = \sqrt{a^2 - b^2} = \sqrt{169 - 144} = \sqrt{25} = 5\]

Thus, the foci are located at (±5, 0). Therefore, the correct answer is option (b) (5, 0) and (-5, 0).
Transcribed Image Text:**Question 23: Find the foci of the ellipse.** \[\frac{x^2}{169} + \frac{y^2}{144} = 1\] Options: - (a) (6, 0) and (-6, 0) - (b) (5, 0) and (-5, 0) - (c) (7, 0) and (-7, 0) - (d) (8, 0) and (-8, 0) **Explanation:** This problem asks to determine the foci of an ellipse given in its standard form equation. The equation of the ellipse is \(\frac{x^2}{169} + \frac{y^2}{144} = 1\), indicating it is a horizontal ellipse because 169 (the larger denominator) is associated with \(x^2\). To find the foci, we use the formula for c, where \(c = \sqrt{a^2 - b^2}\). Here, \(a^2 = 169\) and \(b^2 = 144\). Calculating c: \[c = \sqrt{a^2 - b^2} = \sqrt{169 - 144} = \sqrt{25} = 5\] Thus, the foci are located at (±5, 0). Therefore, the correct answer is option (b) (5, 0) and (-5, 0).
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