0
What else can I help you with?
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?
When you increase the distance between the foci of an ellipse, the eccentricity of the ellipse decreases. The eccentricity of an ellipse is a measure of how elongated or stretched out it is, with a lower eccentricity indicating a more circular shape. So, as you increase the distance between the foci, the shape of the ellipse becomes closer to a circle.
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 can one determine the semi-major axis of an orbit when provided with the perihelion and aphelion distances?
To determine the semi-major axis of an orbit when given the perihelion and aphelion distances, you can use the formula for the semi-major axis, which is the average of the perihelion and aphelion distances. This can be calculated by adding the perihelion and aphelion distances and then dividing by 2.
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.
How do I find the perimeter of an ellipse?
The perimeter of an ellipse cannot be expressed in a simple formula like for a circle. One way to approximate it is by using an elliptic integral, which involves complex mathematical calculations. Alternatively, you can use numerical methods or software to find an accurate approximation of the ellipse's perimeter.
