Graph (x^2)/36+(y^2)/25=1

$\frac{x^{2}}{36} + \frac{y^{2}}{25} = 1$

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 $1$ .

$\frac{x^{2}}{36} + \frac{y^{2}}{25} = 1$

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.

$a = 6$

$b = 5$

$k = 0$

$h = 0$

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.

$\sqrt{{(6)}^{2} - {(5)}^{2}}$

Simplify.

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

$\sqrt{36 - {(5)}^{2}}$

Raise $5$ to the power of $2$ .

$\sqrt{36 - 1 \cdot 25}$

Multiply $- 1$ by $25$ .

$\sqrt{36 - 25}$

Subtract $25$ from $36$ .

$\sqrt{11}$

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.

$(0 + 6, 0)$

Simplify.

$(6, 0)$

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.

$(0 - (6), 0)$

Simplify.

$(- 6, 0)$

Ellipses have two vertices.

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

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

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

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

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.

$(0 + \sqrt{11}, 0)$

Simplify.

$(\sqrt{11}, 0)$

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.

$(0 - (\sqrt{11}), 0)$

Simplify.

$(- \sqrt{11}, 0)$

Ellipses have two foci.

${Focus}_{1}$ : $(\sqrt{11}, 0)$

${Focus}_{2}$ : $(- \sqrt{11}, 0)$

${Focus}_{1}$ : $(\sqrt{11}, 0)$

${Focus}_{2}$ : $(- \sqrt{11}, 0)$

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.

$\frac{\sqrt{{(6)}^{2} - {(5)}^{2}}}{6}$

Simplify the numerator.

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

$\frac{\sqrt{36 - 5^{2}}}{6}$

Raise $5$ to the power of $2$ .

$\frac{\sqrt{36 - 1 \cdot 25}}{6}$

Multiply $- 1$ by $25$ .

$\frac{\sqrt{36 - 25}}{6}$

Subtract $25$ from $36$ .

$\frac{\sqrt{11}}{6}$

Step 9

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

Center: $(0, 0)$

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

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

${Focus}_{1}$ : $(\sqrt{11}, 0)$

${Focus}_{2}$ : $(- \sqrt{11}, 0)$

Eccentricity: $\frac{\sqrt{11}}{6}$

Step 10

Graph (x^2)/36+(y^2)/25=1

Related Questions

Support LCMGCF.com by disabling your adblocker.