Ellipse Equation Analysis

We are given the equation of an ellipse: $3x^2 + 2y^2 = 6$ To find the properties of the ellipse, we first need to convert this equation into its standard form. The standard form of an ellipse centered at the origin is either $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ or $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$, where $a > b$. The major axis is determined by which denominator is larger.

Standardizing the Ellipse Equation

Divide the entire equation by 6 to make the right-hand side equal to 1:

$ \frac{3x^2}{6} + \frac{2y^2}{6} = \frac{6}{6} $

Simplifying this gives us:

$ \frac{x^2}{2} + \frac{y^2}{3} = 1 $

Identifying Major Axis and Vertices

Now, we compare this equation to the standard forms. We have:

• Denominator under $x^2$ is 2.
• Denominator under $y^2$ is 3.

Since the denominator under $y^2$ (which is 3) is greater than the denominator under $x^2$ (which is 2), the major axis lies along the y-axis. In the standard form $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$, we have:

• $a^2 = 3 \implies a = \sqrt{3}$
• $b^2 = 2 \implies b = \sqrt{2}$

The length of the major axis is $2a$. Let's calculate this:

Length of Major Axis = $2a = 2 \times \sqrt{3} = 2\sqrt{3}$

The vertices of an ellipse with the major axis along the y-axis are located at $(0, \pm a)$. Therefore, the coordinates of the vertices are:

Vertices = $(0, \pm \sqrt{3})$

Conclusion

Based on our calculations, the length of the major axis is $2\sqrt{3}$ and the coordinates of the vertices are $(0, \pm \sqrt{3})$.

This matches the second option provided.