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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
This ellipse has a horizontal axis of length 10 and a vertical axis of length 4 What is its equation?
By taking a coordinate system with origin at the center of the ellipse, and x-axis along the major axis, and y-axis along the minor axis, then the ellipse intercepts the x-axis at -5 and 5, and the y-axis at -2 and 2. So that the equation of the ellipse x2/a2 + y2/b2 = 1 becomes x2/52 + y2/22 = 1 or x2/25 + y2/4 = 1.
What is the equation of a circle with radius 10 and center -3 and 6?
Equation of a circle is given by: (x-a)2 + (y-b)2 = r2 Here a & b are the coordinates of the center. So, a = -3 & b = 6. And r = 10. Thus, the equation formed is (x+3)2+(y-6)2 = 102
What are the coordinates for an image on a dilation with a center at the origin?
it is nothing
What are the coordinates of the center of a circle whose diameter has the endpoints (1 4) and (-3 -3)?
Its center is at (-1, 1/2)
The ellipse graphed below has its center at -2 2 its horizontal axis is of length 6 and its vertical axis is of length 10 What is its equation?
23
What are the coordinates of the center of the hyperbola graphed by the equation below?
answer for (x+5)^2/11^2-(y+16)^2/6^2=1 answer for that Question is (-5,-16)
The ellipse graphed below has its center at -14-10 its horizontal axis is of length 12 and its vertical axis is of length 16 What is its equation?
(x+14)2 + (y+10)2 = 1 62 82
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)
What is the equation for an ellipse with center at the origin ,one focus at (1,1) and the length of semi major axise is 4.?
This equation is equal to the first one because it produces the same results, always. ... TL;DR - The circle equation is what you get when you multiply all terms from the ellipse equation by the radius. x^2/a^2 + y^2/b^2 = 1 is an ellipse equation. Well, a circle has a radius where a and b are the same.
What are the coordinates of the center of the circle described by the equation x2 y 52 16?
The equation provided appears to have a typographical error, as it should likely be in the form of a standard circle equation. If you meant (x^2 + y^2 = 16), the center of the circle is at the coordinates (0, 0). If this is not the correct interpretation, please clarify the equation for an accurate response.
How many foci are at the center of an ellipse?
An ellipse has 2 foci. They are inside the ellipse, but they can't be said to be at the centre, as an ellipse doesn't have one.
The equation below describes a circle. What are the coordinates of the center of the circle (x - 6)2 plus (y plus 5)2 152?
The equation of the circle is given by ((x - 6)^2 + (y + 5)^2 = 152). The general form of a circle's equation is ((x - h)^2 + (y - k)^2 = r^2), where ((h, k)) is the center and (r) is the radius. From the equation, the coordinates of the center of the circle are ((6, -5)).
Has its center at 3 5 its horizontal axis is of length 12 and its vertical axis is of length 8 What is its equation?
The equation of the ellipse is [(x-3)/12]2 + [(y-5)/8]2 = 1
How do you write the equation of a circle?
The general equation for the circle - or one of them - is: (x - a)^2 + (y - b)^2 = r^2 Where: a and b are the coordinates of the center r is the radius
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
What is the value of moment of inertia of ellipse about x-Axis?
Moment of inertia about x-axis for an ellipse is = pi*b^3*a /4. Where b is the distance from the center of the ellipse to the outside tip of the minor axis. a is the distance from the ceneter of the ellipse to the outside tip of the major axis. Moment of inertia about x-axis for an ellipse is = pi*b^3*a /4. Where b is the distance from the center of the ellipse to the outside tip of the minor axis. a is the distance from the ceneter of the ellipse to the outside tip of the major axis.
