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What is the equation of an ellipse with vertices 2 0 2 4 and foci 2 1 2 3?
Zachariahwest
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
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What is the standard form equation?
Linear it is Ax + By = C This can be algebraically rearranged to :- By = -Ax + C or y = (-A/B)x + C/B or Ax + By + C = 0 Other conic forms are Parabola Ax^2 + Bx + C = 0 Circle Ax^2 + By^2 + 2Ax + 2By + C = 0 Ellipse x^2 /a^2 + y^2/b^2 = 1
How many more vertices does a quadrilateral have than a triangle?
zero. 0
What is the equation for a circle with a center of (0 0) and a radius of 9?
The equation of the circle is: x^2 + y^2 = 81
Is 2x plus 3 plus 0 a linear equation?
for something to be an equation it has to be equal to something.2x+3=0 would be a linear equation (with x=-1.5) 2x+3+0 is just a series of terms.
What kind of equation is ax plus b equals 0 in?
A linear equation.
In the standard equation for an ellipse a is half the length of the axis?
An ellipse is the set of each and every point in a place such that the sum of the distance from the foci is constant, Major Axis of the ellipse is the part from side to side the center of ellipse to the larger axis, or the length of that sector. The major diameter is the largest diameter of an ellipse. Below equation is the standard ellipse equation: X2/a + Y2/b = 1, (a > b > 0)
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)
Sketch the following 4x2 plus 25y2 equals 100?
4x2 + 25y2 = 100 (divide each element of both sides by 100) x2/25 + y2/4 = 1 This is the equation of an ellipse of the form x2/a2 + y2/b2 = 1, whose center is at (0, 0), a = 5, b = 2, and so c = √(a2 - b2) = √19) Since the major axis is horizontal and it lies on the x-axis, the vertices are 5 units to the left and 5 units to the right of the center (0, 0). Vertices are (-5, 0) and (5, 0). The minor axis is vertical and it lies on the y-axis, so the graph of the ellipse crosses the y-axis at the points (0, -2) and (0, 2), since b = 2) The foci are √19 units to the left and √19 units to the right of the center (0, 0). The foci are (-√19, 0) and (√19, 0).
How does the numerical value of e change the shape of an ellipse?
The eccentricity of an ellipse, e, is the ratio of the distance between the foci to the length of the semi-major axis. As e increases from 0 to 1, the ellipse changes from a circle (e = 0) to form a more flat shape until, at e = 1, it is effectively a straight line.
What are the foci of the ellipse of 9 x squared plus 25 y squared plus 100 y - 125 equals 0?
With the equation of an ellipse in the form (x/a)² + (y/b)² = 1 the axes of the ellipse lie on the x and y axes and the foci are √(a² - b²) along the x axis. 9x² + 25y² + 100y - 125 = 0 → (3x)² + 25(y² + 4y + 4 - 4) = 125 → (3x)² +25(y + 2)² - 100 = 125 → (3x)² +25(y + 2)² = 225 → (3x)²/225 + (y + 2)²/9 = 1 → (x/5)² + ((y+2)/3)² = 1 Thus the foci are √(5² - 3²) = √16 = 4 either side of the y-axis, but the y axis has been shifted up by 2, thus the two foci are (-4, -2) and (4, -2).
How does the shape change with increasing eccentricity?
An ellipse whose eccentricity is zero is a circle. As its eccentricity increases, it becomes more and more elliptical, i.e. its foci move farther apart and it appears more "egg-shaped".
This ellipse is centered at the origin and has a horizontal axis of length 14 and a vertical axis of length 16 What is its equation?
The equation is based on formula (x - h)square / A square + (y-k)square / B square = 1. To apply to the above ellipse the equation would be similar to (x- 0) square/ 14 square + (2014 - 0) square / 16 square.
What shape is it 0-veritces 0-edges 1-face?
Wrong Answer: A shape with no vertices, no edges, and 1 face must be a circle or an ellipse, because if it was three dimensional, then it would have to have at least 4 faces.An ellipse has one edge. And, a circle is an ellipse, so same thing there. A triangle can exist in 3d, and it has only one face. So no, a 3d shape does NOT need to have at least 4 faces. It doesn't even need to have at least 2 faces.A plane is 3d, and it only has one face. It also has 0 edges, and 0 vertices. So, as I stated before Capt cadet erased it with the wrong answer:A plane is the answer to the question.
What has 2 faces and 0 vertices?
Nothing has 2 faces and 0 vertices.
What solid figure has 0 flat surface and 0 vertices?
a sphere has 0 flat serfaces 0 vertices
How many vertices are in a circle?
Zero vertices
What shape has 0faces 0 edges 0 vertices?
A sphere has no faces, edges, or vertices.
