Pre-Algebra Examples

$(\frac{x^{2}}{9}) + (\frac{y^{2}}{25}) = 1$

Step 1

Remove parentheses.

$\frac{x^{2}}{9} + \frac{y^{2}}{25} = 1$

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.

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

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.

$a = 5$

$b = 3$

$k = 0$

$h = 0$

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.

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Step 5.1

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

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

Step 5.2

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

$\sqrt{{(5)}^{2} - {(3)}^{2}}$

Step 5.3

Simplify.

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Step 5.3.1

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

$\sqrt{25 - {(3)}^{2}}$

Step 5.3.2

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

$\sqrt{25 - 1 \cdot 9}$

Step 5.3.3

Multiply $- 1$ by $9$ .

$\sqrt{25 - 9}$

Step 5.3.4

Subtract $9$ from $25$ .

$\sqrt{16}$

Step 5.3.5

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

$\sqrt{4^{2}}$

Step 5.3.6

Pull terms out from under the radical, assuming positive real numbers.

$4$

Step 6

Find the vertices.

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Step 6.1

The first vertex of an ellipse can be found by adding $a$ to $k$ .

$(h, k + a)$

Step 6.2

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

$(0, 0 + 5)$

Step 6.3

Simplify.

$(0, 5)$

Step 6.4

The second vertex of an ellipse can be found by subtracting $a$ from $k$ .

$(h, k - a)$

Step 6.5

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

$(0, 0 - (5))$

Step 6.6

Simplify.

$(0, - 5)$

Step 6.7

Ellipses have two vertices.

${Vertex}_{1}$ : $(0, 5)$

${Vertex}_{2}$ : $(0, - 5)$

${Vertex}_{1}$ : $(0, 5)$

${Vertex}_{2}$ : $(0, - 5)$

Step 7

Find the foci.

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Step 7.1

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

$(h, k + c)$

Step 7.2

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

$(0, 0 + 4)$

Step 7.3

Simplify.

$(0, 4)$

Step 7.4

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

$(h, k - c)$

Step 7.5

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

$(0, 0 - (4))$

Step 7.6

Simplify.

$(0, - 4)$

Step 7.7

Ellipses have two foci.

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

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

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

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

Step 8

Find the eccentricity.

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Step 8.1

Find the eccentricity by using the following formula.

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

Step 8.2

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

$\frac{\sqrt{{(5)}^{2} - {(3)}^{2}}}{5}$

Step 8.3

Simplify the numerator.

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Step 8.3.1

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

$\frac{\sqrt{25 - 3^{2}}}{5}$

Step 8.3.2

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

$\frac{\sqrt{25 - 1 \cdot 9}}{5}$

Step 8.3.3

Multiply $- 1$ by $9$ .

$\frac{\sqrt{25 - 9}}{5}$

Step 8.3.4

Subtract $9$ from $25$ .

$\frac{\sqrt{16}}{5}$

Step 8.3.5

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

$\frac{\sqrt{4^{2}}}{5}$

Step 8.3.6

Pull terms out from under the radical, assuming positive real numbers.

$\frac{4}{5}$

Step 9

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

Center: $(0, 0)$

${Vertex}_{1}$ : $(0, 5)$

${Vertex}_{2}$ : $(0, - 5)$

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

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

Eccentricity: $\frac{4}{5}$

Step 10