Precalculus Examples

$4 x^{2} + 9 y^{2} = 1$

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 $1$ .

$\frac{x^{2}}{\frac{1}{4}} + \frac{y^{2}}{\frac{1}{9}} = 1$

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.

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

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.

$a = \frac{1}{2}$

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

$k = 0$

$h = 0$

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.

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

Step 5.3

Simplify.

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Step 5.3.1

Apply the product rule to $\frac{1}{2}$ .

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

Step 5.3.2

One to any power is one.

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

Step 5.3.3

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

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

Step 5.3.4

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

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

Step 5.3.5

One to any power is one.

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

Step 5.3.6

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

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

Step 5.3.7

To write $\frac{1}{4}$ as a fraction with a common denominator, multiply by $\frac{9}{9}$ .

$\sqrt{\frac{1}{4} \cdot \frac{9}{9} - \frac{1}{9}}$

Step 5.3.8

To write $- \frac{1}{9}$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$\sqrt{\frac{1}{4} \cdot \frac{9}{9} - \frac{1}{9} \cdot \frac{4}{4}}$

Step 5.3.9

Write each expression with a common denominator of $36$ , by multiplying each by an appropriate factor of $1$ .

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Step 5.3.9.1

Multiply $\frac{1}{4}$ by $\frac{9}{9}$ .

$\sqrt{\frac{9}{4 \cdot 9} - \frac{1}{9} \cdot \frac{4}{4}}$

Step 5.3.9.2

Multiply $4$ by $9$ .

$\sqrt{\frac{9}{36} - \frac{1}{9} \cdot \frac{4}{4}}$

Step 5.3.9.3

Multiply $\frac{1}{9}$ by $\frac{4}{4}$ .

$\sqrt{\frac{9}{36} - \frac{4}{9 \cdot 4}}$

Step 5.3.9.4

Multiply $9$ by $4$ .

$\sqrt{\frac{9}{36} - \frac{4}{36}}$

Step 5.3.10

Combine the numerators over the common denominator.

$\sqrt{\frac{9 - 4}{36}}$

Step 5.3.11

Subtract $4$ from $9$ .

$\sqrt{\frac{5}{36}}$

Step 5.3.12

Rewrite $\sqrt{\frac{5}{36}}$ as $\frac{\sqrt{5}}{\sqrt{36}}$ .

$\frac{\sqrt{5}}{\sqrt{36}}$

Step 5.3.13

Simplify the denominator.

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Step 5.3.13.1

Rewrite $36$ as $6^{2}$ .

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

Step 5.3.13.2

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

$\frac{\sqrt{5}}{6}$

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 $h$ .

$(h + a, k)$

Step 6.2

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

$(0 + \frac{1}{2}, 0)$

Step 6.3

Simplify.

$(\frac{1}{2}, 0)$

Step 6.4

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

$(h - a, k)$

Step 6.5

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

$(0 - (\frac{1}{2}), 0)$

Step 6.6

Simplify.

$(- \frac{1}{2}, 0)$

Step 6.7

Ellipses have two vertices.

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

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

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

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

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 $h$ .

$(h + c, k)$

Step 7.2

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

$(0 + \frac{\sqrt{5}}{6}, 0)$

Step 7.3

Simplify.

$(\frac{\sqrt{5}}{6}, 0)$

Step 7.4

The second focus of an ellipse can be found by subtracting $c$ from $h$ .

$(h - c, k)$

Step 7.5

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

$(0 - (\frac{\sqrt{5}}{6}), 0)$

Step 7.6

Simplify.

$(- \frac{\sqrt{5}}{6}, 0)$

Step 7.7

Ellipses have two foci.

${Focus}_{1}$ : $(\frac{\sqrt{5}}{6}, 0)$

${Focus}_{2}$ : $(- \frac{\sqrt{5}}{6}, 0)$

${Focus}_{1}$ : $(\frac{\sqrt{5}}{6}, 0)$

${Focus}_{2}$ : $(- \frac{\sqrt{5}}{6}, 0)$

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.

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

Step 8.3

Simplify.

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Step 8.3.1

Multiply the numerator by the reciprocal of the denominator.

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

Step 8.3.2

Apply the product rule to $\frac{1}{2}$ .

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

Step 8.3.3

One to any power is one.

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

Step 8.3.4

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

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

Step 8.3.5

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

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

Step 8.3.6

One to any power is one.

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

Step 8.3.7

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

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

Step 8.3.8

To write $\frac{1}{4}$ as a fraction with a common denominator, multiply by $\frac{9}{9}$ .

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

Step 8.3.9

To write $- \frac{1}{9}$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

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

Step 8.3.10

Write each expression with a common denominator of $36$ , by multiplying each by an appropriate factor of $1$ .

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Step 8.3.10.1

Multiply $\frac{1}{4}$ by $\frac{9}{9}$ .

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

Step 8.3.10.2

Multiply $4$ by $9$ .

$\sqrt{\frac{9}{36} - \frac{1}{9} \cdot \frac{4}{4}} \cdot 2$

Step 8.3.10.3

Multiply $\frac{1}{9}$ by $\frac{4}{4}$ .

$\sqrt{\frac{9}{36} - \frac{4}{9 \cdot 4}} \cdot 2$

Step 8.3.10.4

Multiply $9$ by $4$ .

$\sqrt{\frac{9}{36} - \frac{4}{36}} \cdot 2$

Step 8.3.11

Combine the numerators over the common denominator.

$\sqrt{\frac{9 - 4}{36}} \cdot 2$

Step 8.3.12

Subtract $4$ from $9$ .

$\sqrt{\frac{5}{36}} \cdot 2$

Step 8.3.13

Rewrite $\sqrt{\frac{5}{36}}$ as $\frac{\sqrt{5}}{\sqrt{36}}$ .

$\frac{\sqrt{5}}{\sqrt{36}} \cdot 2$

Step 8.3.14

Simplify the denominator.

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Step 8.3.14.1

Rewrite $36$ as $6^{2}$ .

$\frac{\sqrt{5}}{\sqrt{6^{2}}} \cdot 2$

Step 8.3.14.2

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

$\frac{\sqrt{5}}{6} \cdot 2$

Step 8.3.15

Cancel the common factor of $2$ .

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Step 8.3.15.1

Factor $2$ out of $6$ .

$\frac{\sqrt{5}}{2 (3)} \cdot 2$

Step 8.3.15.2

Cancel the common factor.

$\frac{\sqrt{5}}{2 \cdot 3} \cdot 2$

Step 8.3.15.3

Rewrite the expression.

$\frac{\sqrt{5}}{3}$

Step 9

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

Center: $(0, 0)$

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

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

${Focus}_{1}$ : $(\frac{\sqrt{5}}{6}, 0)$

${Focus}_{2}$ : $(- \frac{\sqrt{5}}{6}, 0)$

Eccentricity: $\frac{\sqrt{5}}{3}$

Step 10