How do you find the major/minor axis of an ellipse?

Question

Find the length of the major/minor axis for the ellipse defined by this equation.

(x+1)^2/64 + (y+1)^2/36 = 1

Please explain in simple terms.

Answer 1

(x+1)² / 64 + (y+1)² / 36 = 1



The major axis will be related to the largest denominator, 64 where 64 = a²

a = 8, the distance from the center of the ellipse to each end of the major axis.

Hence, the length of the major axis is 2(8) = 16.



The minor axis will be related to the smaller denominator, 36 where 36 = b²

b = 6, the distance from the center of the ellipse to each end of the minor axis.

Hence, the length of the minor axis is 2(6) = 12.



I hope this helps you!

Answer 2

An ellipse has a major axis, and a minor axis. The major axis, is the longer axis, with length = 2a, where 2a is also the sum of the distances from a point on the ellipse to each of the foci.



The minor axis, is the shorter axis, with length 2b, where a^2 = c^2 + b^2.



c is the length from the center of the ellipse to either foci

a is the length from the center of the ellipse to the vertices on the ellipse's major axis

b is the length from the center of the ellipse to the vertices on the ellipse's minor axis



The general equation for an ellipse are:



(x - h)^2/a^2 + (y - k)^2/b^2 = 1, where (h, k) is the center, and where the major axis is parallel to the x axis.

and

(x - h)^2/b^2 + (y - k)^2/a^2 = 1, where the major axis is parallel to the y axis.



In your case, 64 > 36, and since 64 is under the x, then the major axis is parallel to the x axis. Since the center of your ellipse is (-1, -1), the line that is parallel to the x axis and crosses the center is given by y = -1.



Similarly, the minor axis is parallel to the y axis, and is given by the equation x = -1



SIMPLE TERMS

Your ellipse is 2√64 wide and 2√36 tall. Since it's wider than it is tall, then the major axis is horizontal, and the minor axis is vertical. The major and minor axes cross the ellipse at its vertices and its center. To find an equation for these axes, consider the center, and set either y = k, if it's horizontal line, or x = h, if it's a vertical line.

Answer 3

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Well, I'm not sure how you've been taught, but here's how I'd do it. We want to get it into the form (1/(a^2))(x - h)^2 + (1/(b^2))(y-k)^2 = 1 Once in this form, the center, foci and axes are easy to find: center = (h,k) If a > b, major axis is x=h, minor is y=k, and foci are at (h+c, k) and (h-c, k) If b > a, major axis is y=k, minor is x=h, and foci are at (h, k+c) and (h, k-c) where c is found using the equation: c^2 = a^2 - b^2. So to get the equation into the right form, do these steps: Move the 4 to the right. 4x^2 + 9y^2 - 8x + 36y = -4 Factor. 4(x^2 - 2x) + 9(y^2 + 4y) = -4 Use completing the square to get (x^2 - 2x) = (x - 1)^2 - 1 & (y^2 + 4y) = (y + 2)^2 - 4 So substituting these into the original equation, we have 4[(x - 1)^2 - 1] + 9[(y + 2)^2 - 4] = -4 Distribute the 4 and 9: 4(x - 1)^2 - 4 + 9(y + 2)^2 - 36 = -4 Bring the -4 and -36 to the right: 4(x - 1)^2 + 9(y + 2)^2 = 36 Now divide everything by 36: (1/9)(x - 1)^2 + (1/4)(y + 2)^2 = 1 So a^2 = 9, b^2 = 4, which means a = 3, b = 2. The center is at (h,k) = (1,-2) and our major axis is x = 1, minor is y=-2 since a > b. To find c, we have c^2 = 9 - 4 = 5, so c = sqrt(5). So our foci are (h+c, k) = (1+sqrt(5), -2) & (h-c, k) = (1-sqrt(5), -2)

Answer 4

Major Axis Of An Ellipse

Answer 5

RE:

How do you find the major/minor axis of an ellipse?

Find the length of the major/minor axis for the ellipse defined by this equation.



(x+1)^2/64 + (y+1)^2/36 = 1



Please explain in simple terms.