Precalculus Examples

$16 x^{2} + 9 y^{2} = 144$

Step 1

Find the standard form of the ellipse.

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Step 1.1

Divide each term by $144$ to make the right side equal to one.

$\frac{16 x^{2}}{144} + \frac{9 y^{2}}{144} = \frac{144}{144}$

Step 1.2

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 $1$ .

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

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.

$a = 4$

$b = 3$

$k = 0$

$h = 0$

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.

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

Step 5.3

Simplify.

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Step 5.3.1

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

$\sqrt{16 - {(3)}^{2}}$

Step 5.3.2

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

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

Step 5.3.3

Multiply $- 1$ by $9$ .

$\sqrt{16 - 9}$

Step 5.3.4

Subtract $9$ from $16$ .

$\sqrt{7}$

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.

$(0, 0 + 4)$

Step 6.3

Simplify.

$(0, 4)$

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.

$(0, 0 - (4))$

Step 6.6

Simplify.

$(0, - 4)$

Step 6.7

Ellipses have two vertices.

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

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

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

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

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.

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

Step 7.3

Simplify.

$(0, \sqrt{7})$

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.

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

Step 7.6

Simplify.

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

Step 7.7

Ellipses have two foci.

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

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

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

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

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.

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

Step 8.3

Simplify the numerator.

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Step 8.3.1

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

$\frac{\sqrt{16 - 3^{2}}}{4}$

Step 8.3.2

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

$\frac{\sqrt{16 - 1 \cdot 9}}{4}$

Step 8.3.3

Multiply $- 1$ by $9$ .

$\frac{\sqrt{16 - 9}}{4}$

Step 8.3.4

Subtract $9$ from $16$ .

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

Step 9

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

Center: $(0, 0)$

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

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

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

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

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

Step 10