Graph ((x-1)^2)/9+((y-3)^2)/4=1

$\frac{{(x - 1)}^{2}}{9} + \frac{{(y - 3)}^{2}}{4} = 1$

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 $1$ .

$\frac{{(x - 1)}^{2}}{9} + \frac{{(y - 3)}^{2}}{4} = 1$

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.

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

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.

$a = 3$

$b = 2$

$k = 3$

$h = 1$

The center of an ellipse follows the form of $(h, k)$ . Substitute in the values of $h$ and $k$ .

$(1, 3)$

Find $c$ , the distance from the center to a focus.

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Find the distance from the center to a focus of the ellipse by using the following formula.

$\sqrt{a^{2} - b^{2}}$

Substitute the values of $a$ and $b$ in the formula.

$\sqrt{{(3)}^{2} - {(2)}^{2}}$

Simplify.

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Raise $3$ to the power of $2$ .

$\sqrt{9 - {(2)}^{2}}$

Raise $2$ to the power of $2$ .

$\sqrt{9 - 1 \cdot 4}$

Multiply $- 1$ by $4$ .

$\sqrt{9 - 4}$

Subtract $4$ from $9$ .

$\sqrt{5}$

Find the vertices.

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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.

$(1 + 3, 3)$

Simplify.

$(4, 3)$

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.

$(1 - (3), 3)$

Simplify.

$(- 2, 3)$

Ellipses have two vertices.

${Vertex}_{1}$ : $(4, 3)$

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

${Vertex}_{1}$ : $(4, 3)$

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

Find the foci.

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The first focus of an ellipse can be found by adding $c$ to $h$ .

$(h + c, k)$

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

$(1 + \sqrt{5}, 3)$

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

$(h - c, k)$

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

$(1 - (\sqrt{5}), 3)$

Simplify.

$(1 - \sqrt{5}, 3)$

Ellipses have two foci.

${Focus}_{1}$ : $(1 + \sqrt{5}, 3)$

${Focus}_{2}$ : $(1 - \sqrt{5}, 3)$

${Focus}_{1}$ : $(1 + \sqrt{5}, 3)$

${Focus}_{2}$ : $(1 - \sqrt{5}, 3)$

Find the eccentricity.

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Find the eccentricity by using the following formula.

$\frac{\sqrt{a^{2} - b^{2}}}{a}$

Substitute the values of $a$ and $b$ into the formula.

$\frac{\sqrt{{(3)}^{2} - {(2)}^{2}}}{3}$

Simplify the numerator.

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Raise $3$ to the power of $2$ .

$\frac{\sqrt{9 - 2^{2}}}{3}$

Raise $2$ to the power of $2$ .

$\frac{\sqrt{9 - 1 \cdot 4}}{3}$

Multiply $- 1$ by $4$ .

$\frac{\sqrt{9 - 4}}{3}$

Subtract $4$ from $9$ .

$\frac{\sqrt{5}}{3}$

These values represent the important values for graphing and analyzing an ellipse.

Center: $(1, 3)$

${Vertex}_{1}$ : $(4, 3)$

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

${Focus}_{1}$ : $(1 + \sqrt{5}, 3)$

${Focus}_{2}$ : $(1 - \sqrt{5}, 3)$

Eccentricity: $\frac{\sqrt{5}}{3}$

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