Find the Vertices 2x^2+3y^2=6

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

Step 1

Find the standard form of the ellipse.

Divide each term by $6$ to make the right side equal to one.

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

Simplify each term in the equation in order to set the right side equal to $1$ . The standard form of an ellipse or hyperbola requires the right side of the equation be $1$ .

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

Step 2

This is the form of an ellipse. Use this form to determine the values used to find the center along with the major and minor axis of the ellipse.

$\frac{{(x - h)}^{2}}{a^{2}} + \frac{{(y - k)}^{2}}{b^{2}} = 1$

Step 3

Match the values in this ellipse to those of the standard form. The variable $a$ represents the radius of the major axis of the ellipse, $b$ represents the radius of the minor axis of the ellipse, $h$ represents the x-offset from the origin, and $k$ represents the y-offset from the origin.

$a = \sqrt{3}$

$b = \sqrt{2}$

$k = 0$

$h = 0$

Step 4

Find the vertices.

The first vertex of an ellipse can be found by adding $a$ to $h$ .

$(h + a, k)$

Substitute the known values of $h$ , $a$ , and $k$ into the formula.

$(0 + \sqrt{3}, 0)$

Simplify.

$(\sqrt{3}, 0)$

The second vertex of an ellipse can be found by subtracting $a$ from $h$ .

$(h - a, k)$

Substitute the known values of $h$ , $a$ , and $k$ into the formula.

$(0 - (\sqrt{3}), 0)$

Simplify.

$(- \sqrt{3}, 0)$

Ellipses have two vertices.

${Vertex}_{1}$ : $(\sqrt{3}, 0)$

${Vertex}_{2}$ : $(- \sqrt{3}, 0)$

${Vertex}_{1}$ : $(\sqrt{3}, 0)$

${Vertex}_{2}$ : $(- \sqrt{3}, 0)$

Step 5

Find the Vertices 2x^2+3y^2=6

Related Questions

Support LCMGCF.com by disabling your adblocker.