write the equation of the ellipse with foci at (0, -sqrt3) and (0, sqrt 3) and for which 2a=4
i've done this befoe, but with squareroots as endpoints. and if 2a=4, a=2, right? Please help me.
To find the equation of an ellipse with known foci and the length of the major axis, you can use the formula for the distance between a point on the ellipse and the foci.
In this case, the two foci are at (0, -sqrt(3)) and (0, sqrt(3)), and since they lie on the y-axis, the major axis is vertical. The length of the major axis, 2a, is given as 4, which means that a is equal to 2.
The distance between a point (x, y) on the ellipse and one of its foci is given by the formula √((x - h)^2 + (y - k)^2) = 2a, where (h, k) are the coordinates of the center of the ellipse.
Since the center of the ellipse is (0, 0) as it lies on the origin, the equation becomes √(x^2 + y^2) = 2(2).
Now, to get rid of the square root, we square both sides of the equation:
x^2 + y^2 = 16
Therefore, the equation of the ellipse with foci at (0, -sqrt(3)) and (0, sqrt(3)) and for which 2a = 4 is x^2 + y^2 = 16.