Find the Vertices (x^2)/16+(y^2)/32=1

$\frac{x^{2}}{16} + \frac{y^{2}}{32} = 1$

Step 1

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 $1$ .

$\frac{x^{2}}{16} + \frac{y^{2}}{32} = 1$

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.

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

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.

$a = 4 \sqrt{2}$

$b = 4$

$k = 0$

$h = 0$

Step 4

Find the vertices.

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

$(h, k + a)$

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

$(0, 0 + 4 \sqrt{2})$

Simplify.

$(0, 4 \sqrt{2})$

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

$(h, k - a)$

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

$(0, 0 - (4 \sqrt{2}))$

Simplify.

$(0, - 4 \sqrt{2})$

Ellipses have two vertices.

${Vertex}_{1}$ : $(0, 4 \sqrt{2})$

${Vertex}_{2}$ : $(0, - 4 \sqrt{2})$

${Vertex}_{1}$ : $(0, 4 \sqrt{2})$

${Vertex}_{2}$ : $(0, - 4 \sqrt{2})$

Step 5

Find the Vertices (x^2)/16+(y^2)/32=1

Related Questions

Support LCMGCF.com by disabling your adblocker.