Please help-What is the equation of the ellipse with foci (0,6), (0,-6) and co-vertices (2,0),(-2,0)
Please explain the steps because I have 5 to do for homework-Thank you-I'm really stuck on this
I'm thinking this should be
a=?
b=2
c=6
a^2 -2^2 = 6^2
a^2 = 40
so the equation would be x^2/4 + y^2/40 = 1
Is that correct?
That is correct solution.
Could you recheck my answer because I just realized it can't be write-it's not a choice-
could it be x^2/1 +y^2/40 =1 or would it be x^2/1 + y^2/36 = 1
Those were the only two choices
To find the equation of an ellipse given its foci and co-vertices, you can follow these steps:
Step 1: Identify the center of the ellipse
The foci of the ellipse are located along the y-axis at (0, 6) and (0, -6). The center of the ellipse is the midpoint between these two foci, which is at (0, 0).
Step 2: Determine the major and minor axes
The major axis of an ellipse is the segment that passes through the center and the foci, and the minor axis is the segment that connects the center and the co-vertices.
In this case, the distance between the foci is 6 + 6 = 12, so the length of the major axis is 2a = 12, which means a = 6.
The distance between the co-vertices is 2, so the length of the minor axis is 2b = 2, which means b = 1.
Step 3: Write the equation of the ellipse using the center and lengths of the axes
The general equation for an ellipse centered at the origin is: x^2/a^2 + y^2/b^2 = 1.
Plugging in the previously determined values, we can write the equation: x^2/6^2 + y^2/1^2 = 1.
Simplifying the equation: x^2/36 + y^2 = 1.
Therefore, the equation of the ellipse with foci (0, 6), (0, -6) and co-vertices (2, 0), (-2, 0) is:
x^2/36 + y^2 = 1.