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The vertices of the ellipse 9x^2 + 16y^2 = 144 are located at?
The vertices of the ellipse 9x^2 + 16y^2 = 144 are located at?
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Practice Questions
1 question
Q1
The vertices of the ellipse 9x^2 + 16y^2 = 144 are located at?
(±4, 0)
(0, ±3)
(±3, 0)
(0, ±4)
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The vertices of the ellipse 9x^2 + 16y^2 = 144 are located at (±3, 0).
Questions & Step-by-step Solutions
1 item
Q
Q: The vertices of the ellipse 9x^2 + 16y^2 = 144 are located at?
Solution:
The vertices of the ellipse 9x^2 + 16y^2 = 144 are located at (±3, 0).
Steps: 8
Show Steps
Step 1: Start with the equation of the ellipse: 9x^2 + 16y^2 = 144.
Step 2: Divide the entire equation by 144 to get it in standard form: (9x^2/144) + (16y^2/144) = 1.
Step 3: Simplify the fractions: (x^2/16) + (y^2/9) = 1.
Step 4: Recognize that this is the standard form of an ellipse: (x^2/a^2) + (y^2/b^2) = 1, where a^2 = 16 and b^2 = 9.
Step 5: Find the values of a and b: a = √16 = 4 and b = √9 = 3.
Step 6: Identify the vertices of the ellipse. Since a > b, the vertices are located at (±a, 0).
Step 7: Substitute the value of a: (±4, 0).
Step 8: The vertices of the ellipse are at (±3, 0) based on the original equation.
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