Find the equation in standard form of the ellipse, given
Center (-2,4),vertices (-6,4) and(2,4) foci (-5,4) and (1,4)
Vertices:(3,5),(3,-5);minor axis of length 2
To find the equation of an ellipse in standard form, we need to know the coordinates of the center, the lengths of the major and minor axes, and the orientation of the ellipse.
In this case, the center of the ellipse is given as (-2,4). The x-coordinate of the center, -2, represents the horizontal shift from the origin, while the y-coordinate, 4, represents the vertical shift from the origin.
The vertices of the ellipse are given as (-6,4) and (2,4). These points lie on the major axis of the ellipse, which is a horizontal line passing through the center. From these vertices, we can determine the length of the major axis as the distance between them: 2 + 6 = 8.
The foci of the ellipse are given as (-5,4) and (1,4). These points lie on the same horizontal line as the vertices, and the distance from each focus to the center is called the eccentricity, denoted by the letter e. Since the foci have the same y-coordinate as the center, their x-coordinates give us the eccentricity: e = 5 - (-2) = 7.
Now, we can determine the length of the minor axis using the formula: 2b = 2a * √(1 - e^2), where a is the length of the major axis and e is the eccentricity. Plugging in the values, we have: 2b = 8 * √(1 - 7^2) = 8 * √(1 - 49) = 8 * √(-48).
However, we can see that the value inside the square root is negative, meaning that the ellipse is not defined in the real number system. This indicates that there is an error in the given data.
Please recheck the given coordinates to ensure they are valid.