Graph (x^2)/81+(y^2)/49=1

$\frac{x^{2}}{81} + \frac{y^{2}}{49} = 1$

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 $1$ .

$\frac{x^{2}}{81} + \frac{y^{2}}{49} = 1$

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.

$a = 9$

$b = 7$

$k = 0$

$h = 0$

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.

$\sqrt{{(9)}^{2} - {(7)}^{2}}$

Simplify.

Raise $9$ to the power of $2$ .

$\sqrt{81 - {(7)}^{2}}$

Raise $7$ to the power of $2$ .

$\sqrt{81 - 1 \cdot 49}$

Multiply $- 1$ by $49$ .

$\sqrt{81 - 49}$

Subtract $49$ from $81$ .

$\sqrt{32}$

Rewrite $32$ as $4^{2} \cdot 2$ .

Factor $16$ out of $32$ .

$\sqrt{16 (2)}$

Rewrite $16$ as $4^{2}$ .

$\sqrt{4^{2} \cdot 2}$

Pull terms out from under the radical.

$4 \sqrt{2}$

Step 6

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.

$(0 + 9, 0)$

Simplify.

$(9, 0)$

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.

$(0 - (9), 0)$

Simplify.

$(- 9, 0)$

Ellipses have two vertices.

${Vertex}_{1}$ : $(9, 0)$

${Vertex}_{2}$ : $(- 9, 0)$

${Vertex}_{1}$ : $(9, 0)$

${Vertex}_{2}$ : $(- 9, 0)$

Step 7

Find the foci.

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.

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

Simplify.

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

The second focus 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.

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

Simplify.

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

Ellipses have two foci.

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

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

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

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

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.

$\frac{\sqrt{{(9)}^{2} - {(7)}^{2}}}{9}$

Simplify the numerator.

Raise $9$ to the power of $2$ .

$\frac{\sqrt{81 - 7^{2}}}{9}$

Raise $7$ to the power of $2$ .

$\frac{\sqrt{81 - 1 \cdot 49}}{9}$

Multiply $- 1$ by $49$ .

$\frac{\sqrt{81 - 49}}{9}$

Subtract $49$ from $81$ .

$\frac{\sqrt{32}}{9}$

Rewrite $32$ as $4^{2} \cdot 2$ .

Factor $16$ out of $32$ .

$\frac{\sqrt{16 (2)}}{9}$

Rewrite $16$ as $4^{2}$ .

$\frac{\sqrt{4^{2} \cdot 2}}{9}$

Pull terms out from under the radical.

$\frac{4 \sqrt{2}}{9}$

Step 9

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

Center: $(0, 0)$

${Vertex}_{1}$ : $(9, 0)$

${Vertex}_{2}$ : $(- 9, 0)$

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

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

Eccentricity: $\frac{4 \sqrt{2}}{9}$

Step 10

Graph (x^2)/81+(y^2)/49=1

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