Graph (x^2)/9+(y^2)/36=1

$\frac{x^{2}}{9} + \frac{y^{2}}{36} = 1$

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 $1$ .

$\frac{x^{2}}{9} + \frac{y^{2}}{36} = 1$

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.

$a = 6$

$b = 3$

$k = 0$

$h = 0$

Step 4

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

$(0, 0)$

Step 5

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

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.

$\sqrt{{(6)}^{2} - {(3)}^{2}}$

Simplify.

Raise $6$ to the power of $2$ .

$\sqrt{36 - {(3)}^{2}}$

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

$\sqrt{36 - 1 \cdot 9}$

Multiply $- 1$ by $9$ .

$\sqrt{36 - 9}$

Subtract $9$ from $36$ .

$\sqrt{27}$

Rewrite $27$ as $3^{2} \cdot 3$ .

Factor $9$ out of $27$ .

$\sqrt{9 (3)}$

Rewrite $9$ as $3^{2}$ .

$\sqrt{3^{2} \cdot 3}$

Pull terms out from under the radical.

$3 \sqrt{3}$

Step 6

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.

$(0, 0 + 6)$

Simplify.

$(0, 6)$

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.

$(0, 0 - (6))$

Simplify.

$(0, - 6)$

Ellipses have two vertices.

${Vertex}_{1}$ : $(0, 6)$

${Vertex}_{2}$ : $(0, - 6)$

${Vertex}_{1}$ : $(0, 6)$

${Vertex}_{2}$ : $(0, - 6)$

Step 7

Find the foci.

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

$(h, k + c)$

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

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

Simplify.

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

The first focus of an ellipse can be found by subtracting $c$ from $k$ .

$(h, k - c)$

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

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

Simplify.

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

Ellipses have two foci.

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

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

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

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

Step 8

Find the eccentricity.

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.

$\frac{\sqrt{{(6)}^{2} - {(3)}^{2}}}{6}$

Simplify.

Simplify the numerator.

Raise $6$ to the power of $2$ .

$\frac{\sqrt{36 - 3^{2}}}{6}$

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

$\frac{\sqrt{36 - 1 \cdot 9}}{6}$

Multiply $- 1$ by $9$ .

$\frac{\sqrt{36 - 9}}{6}$

Subtract $9$ from $36$ .

$\frac{\sqrt{27}}{6}$

Rewrite $27$ as $3^{2} \cdot 3$ .

Factor $9$ out of $27$ .

$\frac{\sqrt{9 (3)}}{6}$

Rewrite $9$ as $3^{2}$ .

$\frac{\sqrt{3^{2} \cdot 3}}{6}$

Pull terms out from under the radical.

$\frac{3 \sqrt{3}}{6}$

Cancel the common factor of $3$ and $6$ .

Factor $3$ out of $3 \sqrt{3}$ .

$\frac{3 (\sqrt{3})}{6}$

Cancel the common factors.

Factor $3$ out of $6$ .

$\frac{3 \sqrt{3}}{3 \cdot 2}$

Cancel the common factor.

$\frac{3 \sqrt{3}}{3 \cdot 2}$

Rewrite the expression.

$\frac{\sqrt{3}}{2}$

Step 9

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

Center: $(0, 0)$

${Vertex}_{1}$ : $(0, 6)$

${Vertex}_{2}$ : $(0, - 6)$

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

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

Eccentricity: $\frac{\sqrt{3}}{2}$

Step 10

Graph (x^2)/9+(y^2)/36=1

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