Find the equation of the ellipse with vertices at (-1,3) and (5,3)and length of minor 4.
first find the centre, which would be the midpoint of the vertices,
it would be (2,3)
the length of the minor axis is 4, so b = 2
sofar we would would have
(x-2)^2/a^2 + (y-3)^2/4 = 1
but (5,3) lies on it, so
(5-2)^2/a^2 + (3-3)^2/4 = 1
9/a^2 = 1
a^2 = 9
final equation
(x-2)^2/9 + (y-3)^2/4 = 1
thanks!
8(16)
To find the equation of the ellipse, we need to determine its center and the lengths of its major and minor axes.
The vertices of the ellipse are given as (-1,3) and (5,3). Since the vertices lie on the same horizontal line, we can conclude that the major axis of the ellipse is parallel to the x-axis. Therefore, the center of the ellipse lies in the middle of the two vertices.
To find the center, we can average the x-coordinates and the y-coordinates of the two vertices:
Center = ((-1 + 5) / 2, (3 + 3) / 2) = (2, 3).
The length of the minor axis is given as 4. This is the distance between the center and one of the points on the minor axis.
Now we have the following information:
Center: (2, 3)
Length of the major axis: The distance between the x-coordinates of the two vertices, which is 5 - (-1) = 6.
Length of the minor axis: 4.
The equation of an ellipse with center (h, k), major axis length 2a, and minor axis length 2b is given by:
((x - h)^2) / (a^2) + ((y - k)^2) / (b^2) = 1.
Now we can plug in the values into the equation:
((x - 2)^2) / (6/2)^2 + ((y - 3)^2) / (4/2)^2 = 1.
Simplifying further:
(x - 2)^2 / 9 + (y - 3)^2 / 4 = 1.
Therefore, the equation of the ellipse is (x - 2)^2 / 9 + (y - 3)^2 / 4 = 1.