Pre-Algebra Examples

$16 x^{2} + 9 y^{2} = 1$

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 $1$ .

$\frac{x^{2}}{\frac{1}{16}} + \frac{y^{2}}{\frac{1}{9}} = 1$

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.

$a = \frac{1}{3}$

$b = \frac{1}{4}$

$k = 0$

$h = 0$

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.

$\sqrt{{(\frac{1}{3})}^{2} - {(\frac{1}{4})}^{2}}$

Step 5.3

Simplify.

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Step 5.3.1

Apply the product rule to $\frac{1}{3}$ .

$\sqrt{\frac{1^{2}}{3^{2}} - {(\frac{1}{4})}^{2}}$

Step 5.3.2

One to any power is one.

$\sqrt{\frac{1}{3^{2}} - {(\frac{1}{4})}^{2}}$

Step 5.3.3

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

$\sqrt{\frac{1}{9} - {(\frac{1}{4})}^{2}}$

Step 5.3.4

Apply the product rule to $\frac{1}{4}$ .

$\sqrt{\frac{1}{9} - \frac{1^{2}}{4^{2}}}$

Step 5.3.5

One to any power is one.

$\sqrt{\frac{1}{9} - \frac{1}{4^{2}}}$

Step 5.3.6

Raise $4$ to the power of $2$ .

$\sqrt{\frac{1}{9} - \frac{1}{16}}$

Step 5.3.7

To write $\frac{1}{9}$ as a fraction with a common denominator, multiply by $\frac{16}{16}$ .

$\sqrt{\frac{1}{9} \cdot \frac{16}{16} - \frac{1}{16}}$

Step 5.3.8

To write $- \frac{1}{16}$ as a fraction with a common denominator, multiply by $\frac{9}{9}$ .

$\sqrt{\frac{1}{9} \cdot \frac{16}{16} - \frac{1}{16} \cdot \frac{9}{9}}$

Step 5.3.9

Write each expression with a common denominator of $144$ , by multiplying each by an appropriate factor of $1$ .

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Step 5.3.9.1

Multiply $\frac{1}{9}$ by $\frac{16}{16}$ .

$\sqrt{\frac{16}{9 \cdot 16} - \frac{1}{16} \cdot \frac{9}{9}}$

Step 5.3.9.2

Multiply $9$ by $16$ .

$\sqrt{\frac{16}{144} - \frac{1}{16} \cdot \frac{9}{9}}$

Step 5.3.9.3

Multiply $\frac{1}{16}$ by $\frac{9}{9}$ .

$\sqrt{\frac{16}{144} - \frac{9}{16 \cdot 9}}$

Step 5.3.9.4

Multiply $16$ by $9$ .

$\sqrt{\frac{16}{144} - \frac{9}{144}}$

Step 5.3.10

Combine the numerators over the common denominator.

$\sqrt{\frac{16 - 9}{144}}$

Step 5.3.11

Subtract $9$ from $16$ .

$\sqrt{\frac{7}{144}}$

Step 5.3.12

Rewrite $\sqrt{\frac{7}{144}}$ as $\frac{\sqrt{7}}{\sqrt{144}}$ .

$\frac{\sqrt{7}}{\sqrt{144}}$

Step 5.3.13

Simplify the denominator.

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Step 5.3.13.1

Rewrite $144$ as $12^{2}$ .

$\frac{\sqrt{7}}{\sqrt{12^{2}}}$

Step 5.3.13.2

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

$\frac{\sqrt{7}}{12}$

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.

$(0, 0 + \frac{1}{3})$

Step 6.3

Simplify.

$(0, \frac{1}{3})$

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.

$(0, 0 - (\frac{1}{3}))$

Step 6.6

Simplify.

$(0, - \frac{1}{3})$

Step 6.7

Ellipses have two vertices.

${Vertex}_{1}$ : $(0, \frac{1}{3})$

${Vertex}_{2}$ : $(0, - \frac{1}{3})$

${Vertex}_{1}$ : $(0, \frac{1}{3})$

${Vertex}_{2}$ : $(0, - \frac{1}{3})$

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.

$(0, 0 + \frac{\sqrt{7}}{12})$

Step 7.3

Simplify.

$(0, \frac{\sqrt{7}}{12})$

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.

$(0, 0 - (\frac{\sqrt{7}}{12}))$

Step 7.6

Simplify.

$(0, - \frac{\sqrt{7}}{12})$

Step 7.7

Ellipses have two foci.

${Focus}_{1}$ : $(0, \frac{\sqrt{7}}{12})$

${Focus}_{2}$ : $(0, - \frac{\sqrt{7}}{12})$

${Focus}_{1}$ : $(0, \frac{\sqrt{7}}{12})$

${Focus}_{2}$ : $(0, - \frac{\sqrt{7}}{12})$

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.

$\frac{\sqrt{{(\frac{1}{3})}^{2} - {(\frac{1}{4})}^{2}}}{\frac{1}{3}}$

Step 8.3

Simplify.

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Step 8.3.1

Multiply the numerator by the reciprocal of the denominator.

$\sqrt{{(\frac{1}{3})}^{2} - {(\frac{1}{4})}^{2}} \cdot 3$

Step 8.3.2

Apply the product rule to $\frac{1}{3}$ .

$\sqrt{\frac{1^{2}}{3^{2}} - {(\frac{1}{4})}^{2}} \cdot 3$

Step 8.3.3

One to any power is one.

$\sqrt{\frac{1}{3^{2}} - {(\frac{1}{4})}^{2}} \cdot 3$

Step 8.3.4

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

$\sqrt{\frac{1}{9} - {(\frac{1}{4})}^{2}} \cdot 3$

Step 8.3.5

Apply the product rule to $\frac{1}{4}$ .

$\sqrt{\frac{1}{9} - \frac{1^{2}}{4^{2}}} \cdot 3$

Step 8.3.6

One to any power is one.

$\sqrt{\frac{1}{9} - \frac{1}{4^{2}}} \cdot 3$

Step 8.3.7

Raise $4$ to the power of $2$ .

$\sqrt{\frac{1}{9} - \frac{1}{16}} \cdot 3$

Step 8.3.8

To write $\frac{1}{9}$ as a fraction with a common denominator, multiply by $\frac{16}{16}$ .

$\sqrt{\frac{1}{9} \cdot \frac{16}{16} - \frac{1}{16}} \cdot 3$

Step 8.3.9

To write $- \frac{1}{16}$ as a fraction with a common denominator, multiply by $\frac{9}{9}$ .

$\sqrt{\frac{1}{9} \cdot \frac{16}{16} - \frac{1}{16} \cdot \frac{9}{9}} \cdot 3$

Step 8.3.10

Write each expression with a common denominator of $144$ , by multiplying each by an appropriate factor of $1$ .

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Step 8.3.10.1

Multiply $\frac{1}{9}$ by $\frac{16}{16}$ .

$\sqrt{\frac{16}{9 \cdot 16} - \frac{1}{16} \cdot \frac{9}{9}} \cdot 3$

Step 8.3.10.2

Multiply $9$ by $16$ .

$\sqrt{\frac{16}{144} - \frac{1}{16} \cdot \frac{9}{9}} \cdot 3$

Step 8.3.10.3

Multiply $\frac{1}{16}$ by $\frac{9}{9}$ .

$\sqrt{\frac{16}{144} - \frac{9}{16 \cdot 9}} \cdot 3$

Step 8.3.10.4

Multiply $16$ by $9$ .

$\sqrt{\frac{16}{144} - \frac{9}{144}} \cdot 3$

Step 8.3.11

Combine the numerators over the common denominator.

$\sqrt{\frac{16 - 9}{144}} \cdot 3$

Step 8.3.12

Subtract $9$ from $16$ .

$\sqrt{\frac{7}{144}} \cdot 3$

Step 8.3.13

Rewrite $\sqrt{\frac{7}{144}}$ as $\frac{\sqrt{7}}{\sqrt{144}}$ .

$\frac{\sqrt{7}}{\sqrt{144}} \cdot 3$

Step 8.3.14

Simplify the denominator.

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Step 8.3.14.1

Rewrite $144$ as $12^{2}$ .

$\frac{\sqrt{7}}{\sqrt{12^{2}}} \cdot 3$

Step 8.3.14.2

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

$\frac{\sqrt{7}}{12} \cdot 3$

Step 8.3.15

Cancel the common factor of $3$ .

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Step 8.3.15.1

Factor $3$ out of $12$ .

$\frac{\sqrt{7}}{3 (4)} \cdot 3$

Step 8.3.15.2

Cancel the common factor.

$\frac{\sqrt{7}}{3 \cdot 4} \cdot 3$

Step 8.3.15.3

Rewrite the expression.

$\frac{\sqrt{7}}{4}$

Step 9

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

Center: $(0, 0)$

${Vertex}_{1}$ : $(0, \frac{1}{3})$

${Vertex}_{2}$ : $(0, - \frac{1}{3})$

${Focus}_{1}$ : $(0, \frac{\sqrt{7}}{12})$

${Focus}_{2}$ : $(0, - \frac{\sqrt{7}}{12})$

Eccentricity: $\frac{\sqrt{7}}{4}$

Step 10