Write the standard equation of the ellipse with the given characteristics
this is due in the morning I have no ideal how to do it
vertices:(-8,1)(0,1) (-4,7) (-4,5)
center:(-4,1)
Thank you for helping me
As I stated in the previous post to this question, you have a typo and the way it is written, we cannot do this question.
But I have a feeling that the last point should be (-4,-5), that way (-4,1) is the midpoint of both axes
so 2a = 8, ----> a = 4
2b = 12 , ----> b = 6
equation:
(x+4)^2 /16 + (y-1)^2 /36 = 1
To find the standard equation of an ellipse given its characteristics, you can use the following formula:
(x - h)^2 / a^2 + (y - k)^2 / b^2 = 1
where (h, k) represents the center of the ellipse, and a and b are the lengths of the semi-major and semi-minor axes, respectively.
In this case, the given center of the ellipse is (-4, 1). To find the values of a and b, we can use the distance formula between the center and the vertices.
1) Distance between the center and vertex along the x-axis:
The x-coordinate of the center is -4, and the x-coordinate of one of the vertices is -8. Using the distance formula:
a = |-4 - (-8)| = 4
2) Distance between the center and vertex along the y-axis:
The y-coordinate of the center is 1, and the y-coordinate of one of the vertices is 1. Using the distance formula:
b = |1 - 1| = 0 (since the distance between them is 0)
You should note that when the distance between the center and vertex along the y-axis is 0, it means the ellipse is actually degenerated into a line segment.
Now that we have the values of a and b, we can substitute them into the equation to obtain the standard form of the ellipse:
(x - (-4))^2 / 4^2 + (y - 1)^2 / 0^2 = 1
Simplifying further:
(x + 4)^2 / 16 + (y - 1)^2 / 0 = 1
Now, we have the standard equation of the ellipse with the given characteristics:
(x + 4)^2 / 16 + (y - 1)^2 = 1
Remember to revise the equation and check if all the given values and calculations are accurate before submitting your work.