Find the vertices of the ellipse x^2/36 + y^2/12 = 1
- Thread starter norton
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I'm sorry, but I don't see where you "worked" anything...?norton said:
Your (somewhat cryptic) reply would seem to imply that you did some sort of graph. Does your graphing utility do graphs of "(y<sup>2</sup> / 36) =", rather than the usual "y="?
Please clarify. When you reply, please show how far you have gotten in the suggested solution process of testing each answer option.
Thank you.
Eliz.
The general form for the equation of an ellipse is
(x - h)^2 / (a^2) + (y - k)^2 / b^2 = 2
The center of the ellipse is at (h, k)
Your equation is
x^2 / 36 + y^2 / 12 = 1
Ok.....since (x - h)^2 is x^2, then h = 0. And, since (y - k)^2 = y^2, k = 0.
The center of your ellipse is (0, 0).
I would HOPE that you know a few things about what the equation of an ellipse tells you about the ellipse.
If the denominator of the fraction containing x^2 is larger than the denominator of the fraction containing y^2, then the "major axis" of the ellipse is horizontal, and the vertices of the ellipse lie on the horizantal line through the center. The distance from the center to each vertex is the square root of the denominator of the fraction with x^2 as the numerator.
If the denominator of the fraction containing y^2 is larger, then the major axis of the ellipse is on the vertical line through the center. The distance from the center of the ellipse to each vertex is the square root of the denominator of the fraction with y^2 as its numerator.
The vertices of the ellipse lie on the major axis.
The distance from the center to the vertex of the ellipse is the square root of
x^2 / 36 + y^2 / 12 = 1
The center of the ellipse is at (0, 0).
The major axis of the ellipse lies on a horizontal line through (0, 0). The distance from the center (0, 0) is sqrt(36)...or 6.
OK...center is at (0, 0). The vertices should be 6 units to the right of the center, and 6 units to the left of the center.
Can you pick the correct answer choice?
(x - h)^2 / (a^2) + (y - k)^2 / b^2 = 2
The center of the ellipse is at (h, k)
Your equation is
x^2 / 36 + y^2 / 12 = 1
Ok.....since (x - h)^2 is x^2, then h = 0. And, since (y - k)^2 = y^2, k = 0.
The center of your ellipse is (0, 0).
I would HOPE that you know a few things about what the equation of an ellipse tells you about the ellipse.
If the denominator of the fraction containing x^2 is larger than the denominator of the fraction containing y^2, then the "major axis" of the ellipse is horizontal, and the vertices of the ellipse lie on the horizantal line through the center. The distance from the center to each vertex is the square root of the denominator of the fraction with x^2 as the numerator.
If the denominator of the fraction containing y^2 is larger, then the major axis of the ellipse is on the vertical line through the center. The distance from the center of the ellipse to each vertex is the square root of the denominator of the fraction with y^2 as its numerator.
The vertices of the ellipse lie on the major axis.
The distance from the center to the vertex of the ellipse is the square root of
x^2 / 36 + y^2 / 12 = 1
The center of the ellipse is at (0, 0).
The major axis of the ellipse lies on a horizontal line through (0, 0). The distance from the center (0, 0) is sqrt(36)...or 6.
OK...center is at (0, 0). The vertices should be 6 units to the right of the center, and 6 units to the left of the center.
Can you pick the correct answer choice?