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How do you find foci of ellipse?
Let's start with the equation of the ellipse. x2/a2 +y2 /b2 =1 This ellipse is centered at the origin, and we can move it by subtracting h from x and k from y and then squaring that quantity. For example, if we move it h units horizontally, we have (x-h)2 instead of just x2 . In any case. b2 =a2 -c2 . The foci are located 2c units part. So if it is centered at the origin, we can just find 2c and each focus is at + or - c. If we move the ellipse, we can still do the same thing, we just need to take into account how much we moved it. Here is an example to help you see it. Vertices (4,0) and (-4,0) center (0,0) End points of minor axis (0,2) and (0,-2) Foci at (3.5,0) and (-3.5,0)
How do you find the surface area of an ellipse?
An ellipse is a two dimensional shape, so it does not have a "surface area", only an "area". Any ellipse has two radii, the major one and the minor one. We'll call them R1 and R2. The area of the ellipse then can be calculated with the function: a = πR1R2 You will notice that this is the same equation as the area for a circle. The circle is a special case though, because it is an ellipse in which both axes are the same length. In that case, R1 equals R2, so we can simply call it r and say: a = πr2
How do you find the x intercept for x2 36 plus y2 64 equals 1?
It seems that it is the equation of an ellipse, x2/b2 + y2/a2 = 1, with center at the origin, a vertical major axis with length 2a, and a horizontal minor axis with length 2b. x2/36 + y2/64 = 1 Since you are looking for the x-intercepts, they are (√b2, 0) = (√36, 0) = (6, 0) and (-√b2, 0) = (-√36, 0) = (-6, 0)
How do you find the circumference of 9?
A circle or ellipse has a circumference. A number, such as 9, does not.
Do you find the maximum value on the x or y-axis?
On the y-axis, normally.
How do you find the major axis in an ellipse?
The major axis is the line that joins the two foci (focuses) of the ellipse. If all you have is a picture of an ellipse and you don't know where the foci are, you can still find the major axis in a few seconds: It's the longest possible line that you can draw completely inside the ellipse, and it's the line straight across the ellipse between the two opposite "points of the egg".
How do you find the center and semi-major and semi-minor axis for ellipse with equation 16x2 plus y2 equals 16?
An ellipse with centre (xo, yo) with major and minor axes a and b (the larger of a, b being the major axis) has an equation of the form: (x - xo)2 / a2 + (y - yo)2 / b2 = 1 The semi-major and semi-minor axes are half the major and minor axes. So re-arrange the equation into this form: 16x2 + y2 = 16 x2 + y2 / 16 = 1 (x - 0)2 / 12 + (y - 0)2 / 42 = 1 Giving: Centre = (0, 0) Major axis = 2 Semi-major axis = 2/2 = 1 Minor axis = 1 Semi-minor axis = 1/2
What changes takes place in the eccentry of the ellipses when you increase the distance between foci?
"You can see this in practical terms by making an ellipse yourself. Put two thumbtacks about four inches apart in a paper-sized piece of cardboard. (These distances are arbitrary, but it will get you started.) Next, tie a loose-fitting string, about seven inches long or so, between the two thumbtacks. Finally, place a pencil inside the string and pull it away from the tacks until the string is taut. The string will look like a triangle with the pencil and the two tacks in the corners. Move the pencil all the way around the cardboard, keeping the string taut, and you will draw an ellipse. This is illustrated in Figure 2. For the purposes of this article, "F1" will be the left focus point for a horizontal ellipse, or the top focus point for a vertical ellipse. Similarly, "F2" will be the right or bottom focus point, for horizontal or vertical ellipses, respectively. Now that you've drawn your ellipse, you can move from arts and crafts to astronomy. In 1609, Johannes Kepler reasoned that the planetary orbits were not circular as once thought, but were elliptical with the Sun at one of the two focus points of the elliptical orbit. So, to simulate an orbit, you must be able to determine the location of the foci, and use one as the Sun. Calculating the Foci To calculate the focus points, you need to know a few basic things. Using correct terminology, the longest axis, going through the two focus points, is called the major axis. The axis perpendicular to this axis at the center of the ellipse is called the minor axis. Half the major axis, marked in Figure 2 by a, is called the semimajor axis. Half the minor axis, indicated as b in Figure 2, is called the semiminor axis. Often, the first time learning a subject, it's helpful to work with terms you are comfortable with. To simplify, I'll use the terms "width" and "height," and "xRadius" and "yRadius" when discussing the major, minor, semimajor, and semiminor axes, respectively. There's still one more thing you have to figure out before you can continue. To find the focus points of an ellipse, you must calculate the eccentricity of the ellipse. This is how elongated it is. Once you know the eccentricity of an ellipse, you can multiply this factor by the "xRadius" (or a in Figure 2) to get the distance from the center point to a focus point. This is marked by ae in Figure 2 and is the offset distance you'll use in your script to move the ellipse to the correct new location. There are a few ways to calculate the eccentricity of an ellipse. The formula I'll use is: (If you're interested in knowing how this formula was derived, check out the Related Resources on ellipses, left column.) Writing that in ActionScript, using the aforementioned simplified terms, the formula becomes: e = Math.sqrt(1 - (yRadius*yRadius) / (xRadius*xRadius)) where Math.sqrt() is the Math object notation for square root. Once you have the eccentricity of the ellipse, all you have to do is multiply that by the "xRadius" (in the case of our horizontal ellipse) to get the distance from the center of the ellipse to one of the focus points around which you can orbit your MovieClip. You may wish to orbit a MovieClip around a specific point, or it might even be the location of another MovieClip. For example, you might again want your example to show the Earth orbiting the Sun and the Moon orbiting the Earth. But which focus point should you use? Since the ae (or xRadius * e) distance you just calculated is the same from the center to each focus point, the last step you need to take is to determine how to apply the offset. One method would be to use a conditional statement (if or switch, for example) to specify whether you add or subtract the offset from the center point of your ellipse. But there is a simpler way. If you use -1 to represent the left/top focus, 0 to represent the center, and 1 to represent the right/bottom focus, you can multiply the ae offset by this factor and subtract it from the desired anchor point. For example, say you are placing your ellipse in the center of your stage at (275,200); you have an ellipse that is 250 pixels wide and 150 pixels tall. This results in an eccentricity of .8. Since your ellipse is 250 pixels wide, the "xRadius" is 125, or half the width. Therefore, the x offset would be calculated like this: centerX -= (xRadius * e) * ellipseFocusPoint; 1) -1 for left focus point 275 -= (125*.8) * -1 275 - (-100)" - alex from yahoo answers
Where can one find information about Ellipse products?
Someone wanting to find information about the Ellipse products can find it online. Someone can look at the adobe site, as well as the Ellipse site itself to find all the information needed.
How do you find volume of ellipse?
An ellipse is a 2-dimensional object and so its volume must be zero!
How do you find foci of ellipse?
Let's start with the equation of the ellipse. x2/a2 +y2 /b2 =1 This ellipse is centered at the origin, and we can move it by subtracting h from x and k from y and then squaring that quantity. For example, if we move it h units horizontally, we have (x-h)2 instead of just x2 . In any case. b2 =a2 -c2 . The foci are located 2c units part. So if it is centered at the origin, we can just find 2c and each focus is at + or - c. If we move the ellipse, we can still do the same thing, we just need to take into account how much we moved it. Here is an example to help you see it. Vertices (4,0) and (-4,0) center (0,0) End points of minor axis (0,2) and (0,-2) Foci at (3.5,0) and (-3.5,0)
How do you find the surface area of an ellipse?
An ellipse is a two dimensional shape, so it does not have a "surface area", only an "area". Any ellipse has two radii, the major one and the minor one. We'll call them R1 and R2. The area of the ellipse then can be calculated with the function: a = πR1R2 You will notice that this is the same equation as the area for a circle. The circle is a special case though, because it is an ellipse in which both axes are the same length. In that case, R1 equals R2, so we can simply call it r and say: a = πr2
What is the equation of an ellipse with vertices 2 0 2 4 and foci 2 1 2 3?
Vertices and the foci lie on the line x =2 Major axis is parellel to the y-axis b > a Center of the ellipse is the midpoint (h,k) of the vertices (2,2) Equation of the ellipse is (x - (2) )^2 / a^2 + (y - (2) )^2 / b^2 Equation of the ellipse is (x-2)^2 / a^2 + (y-2)^2 / b^2 The distance between the center and one of the vertices is b The distance between(2,2) and (2,4) is 2, so b = 2 The distance between the center and one of the foci is c The distance between(2,2) and (2,1) is 1, so c = 1 Now that we know b and c, we can find a^2 c^2=b^2-a^2 (1)^2=(2)^2-a^2 a^2 = 3 The equation of the ellipse is Equation of the ellipse is (x-2)^2 / 3 + (y-2)^2 / 4 =1
How do you find the x intercept for x2 36 plus y2 64 equals 1?
It seems that it is the equation of an ellipse, x2/b2 + y2/a2 = 1, with center at the origin, a vertical major axis with length 2a, and a horizontal minor axis with length 2b. x2/36 + y2/64 = 1 Since you are looking for the x-intercepts, they are (√b2, 0) = (√36, 0) = (6, 0) and (-√b2, 0) = (-√36, 0) = (-6, 0)
How do you find the circumference of 9?
A circle or ellipse has a circumference. A number, such as 9, does not.
What is the semi minor axis length of earths elliptic orbit?
The equation to find the semi-minor axis of elliptical orbit is b=a*sqrt(1 - e^2), where b is the semi-minor axis, a is the semi-major axis, and e is the eccentricity. Therefore, using 17.8AU as the semi-major axis and 0.967 as the eccentricity, the semi-minor axis is calculated to be 4.53AU or 6.62*10^11 m.
How do you find the x-axis on a line graph?
The x-axis is the horizontal line that correlates with the vertical y-axis.
