What are the co-vertices of the ellipse 3x^2+2y^2=72?
thank you....
divide by 72
x^2/24 + y^2/36 = 1
the semi-major axes have lengths √24 and 6
The center is at (0,0), so the candidates are
(±√24,0) and (0,±6)
Now, which of these are the vertices?
Is it (0,±6)?
To find the co-vertices of an ellipse, we need to identify the major and minor axes first.
The given equation of the ellipse is 3x^2 + 2y^2 = 72.
To bring the equation to the standard form, we divide both sides by 72 to get:
(x^2 / 24) + (y^2 / 36) = 1.
We can see that the equation has a larger coefficient for y^2, which means the major axis is vertical and the minor axis is horizontal.
The formula for the co-vertices of an ellipse is given by (±a, 0) for a vertical ellipse, and (0, ±b) for a horizontal ellipse, where a represents the semi-major axis and b represents the semi-minor axis.
From the given equation, we can determine the values of a and b:
a = sqrt(36) = 6 (semi-major axis)
b = sqrt(24) ≈ 4.899 (semi-minor axis)
Since the major axis is vertical, the co-vertices will be located at (0, ±b):
Co-vertices: (0, ±4.899)
To find the co-vertices of an ellipse, we need to determine the length of the minor axis.
The standard form of an ellipse is given by the equation:
((x-h)^2)/a^2 + ((y-k)^2)/b^2 = 1
where (h, k) represents the center of the ellipse, "a" represents the length of the semi-major axis, and "b" represents the length of the semi-minor axis.
In the given equation, 3x^2 + 2y^2 = 72, we need to put it in the standard form.
First, divide both sides of the equation by 72 to make the right side equal to 1:
(3x^2)/72 + (2y^2)/72 = 1
Simplifying, we have:
x^2/24 + y^2/36 = 1
Now, we can see that "a^2" corresponds to 36, and "b^2" corresponds to 24.
The length of the semi-minor axis, "b," is the square root of 24, which is approximately 4.898.
So, the co-vertices would be located at a distance of 4.898 units from the center of the ellipse, horizontally.
Therefore, the co-vertices are (-4.898, 0) and (4.898, 0).