Find the equation of the ellipse whose semi-major axis has length 6 and whose foci are at (3,-2+/-sqrt11).
I have the focal constant = 2a which means that a=3 but don't know how to proceed.
Thanks.
Well you know the center is halfway between the foci, at (3,-2)
I agree that a = 3
The form is:
(y+2)^2/a^2 + (x-3)^2/b^2 = 1
we need b
the distance from center to focus is sqrt(a^2-b^2)
so
11 = 3^2+b^2
b^2 = 2
b = sqrt 2
so
(y+2)^2/9 + (x-3)^2/2 = 1
To find the equation of the ellipse, you can use the standard form equation for an ellipse:
(x-h)^2/a^2 + (y-k)^2/b^2 = 1
where (h, k) is the center of the ellipse, a is the semi-major axis length, and b is the semi-minor axis length.
In this case, you already found that a=3, which is the length of the semi-major axis. To find the center of the ellipse, you can take the average of the foci coordinates. The center is given by (h, k), so we have:
h = (3 + 3) / 2 = 3
k = (-2 + sqrt(11) - 2 - sqrt(11)) / 2 = 0
Therefore, the center of the ellipse is (3, 0).
The distance between the center and each of the foci is given by the semi-major axis length, which is 3. So, the distance between (3, 0) and each of the foci is 3.
Now, let's find the coordinates of the foci. You are given that the foci are at (3, -2 +/- sqrt(11)).
The coordinates of the foci are (h, k +/- c), where c is the distance from the center to each focus. In this case, c = 3.
Thus, the coordinates of the foci are:
(3, -2 + sqrt(11)) and (3, -2 - sqrt(11)).
Now, you have the center (h, k), which is (3, 0), and the coordinates of the foci. Substitute these values into the standard form equation for an ellipse:
(x-3)^2/3^2 + (y-0)^2/b^2 = 1
Simplifying the equation, we get:
(x-3)^2/9 + y^2/b^2 = 1
Since the focus points lie on the vertical axis, the equation of the ellipse has a vertical major axis. Therefore, b is the length of the semi-minor axis.
You are not given the value of b in this problem, so you would need more information to find it and fully determine the equation of the ellipse.