What is the equation of an ellipse with a vertex of (0, 3) and a co-vertex of (-2 , 0)?
i think its 9x^2+4y^2=36...not sure
The equation in standard form:
x^2 / 4 + y^2 / 9 = 1
Therefore, you are correct! Good job!
Thank you :-)
To find the equation of an ellipse, we need to determine the values for a^2 and b^2, where a is the distance from the center to the vertices, and b is the distance from the center to the co-vertices. Let's use the given information to calculate these values.
The vertex is (0, 3), which means the distance from the center to the vertex is a = 3.
The co-vertex is (-2, 0), which means the distance from the center to the co-vertex is b = 2.
The equation of an ellipse centered at the origin is given by:
x^2/a^2 + y^2/b^2 = 1
Substituting the values of a and b, we get:
x^2/3^2 + y^2/2^2 = 1
Simplifying further, we have:
x^2/9 + y^2/4 = 1
Therefore, the correct equation for the ellipse is 9x^2 + 4y^2 = 36.
To find the equation of an ellipse, you need to have information about the center of the ellipse, its major and minor axes, and the orientation of the ellipse.
Given that the vertex is (0, 3) and the co-vertex is (-2, 0), we can deduce that the center of the ellipse is at (h, k), where h is the x-coordinate of the vertex (0) and k is the y-coordinate of the co-vertex (0). Therefore, the center of the ellipse is at (0, 0).
The major axis of the ellipse is the distance between the two vertices, which in this case is 2a units. So, the major axis of the ellipse is 2a = 2(3) = 6 units.
The minor axis of the ellipse is the distance between the two co-vertices, which in this case is 2b units. So, the minor axis of the ellipse is 2b = 2(-2) = -4 units.
Now, we can proceed to find the equation of the ellipse using the formula:
(x-h)^2/a^2 + (y-k)^2/b^2 = 1
Substituting the values we have:
(x-0)^2/3^2 + (y-0)^2/(-2)^2 = 1
Simplifying this equation further:
x^2/9 + y^2/4 = 1
So, the equation of the ellipse with a vertex of (0, 3) and a co-vertex of (-2, 0) is indeed 9x^2 + 4y^2 = 36. Good job!