Find the Vertices ((x+2)^2)/16+((y-1)^2)/9=1

$\frac{{(x + 2)}^{2}}{16} + \frac{{(y - 1)}^{2}}{9} = 1$

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 $1$ .

$\frac{{(x + 2)}^{2}}{16} + \frac{{(y - 1)}^{2}}{9} = 1$

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.

$a = 4$

$b = 3$

$k = 1$

$h = - 2$

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.

$(- 2 + 4, 1)$

Simplify.

$(2, 1)$

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.

$(- 2 - (4), 1)$

Simplify.

$(- 6, 1)$

Ellipses have two vertices.

${Vertex}_{1}$ : $(2, 1)$

${Vertex}_{2}$ : $(- 6, 1)$

${Vertex}_{1}$ : $(2, 1)$

${Vertex}_{2}$ : $(- 6, 1)$

Step 5

Find the Vertices ((x+2)^2)/16+((y-1)^2)/9=1

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