Precalculus Examples

$x^{2} + 9 y^{2} + 2 x + 18 y - 26 = 0$

Step 1

Find the standard form of the ellipse.

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Step 1.1

Add $26$ to both sides of the equation.

$x^{2} + 9 y^{2} + 2 x + 18 y = 26$

Step 1.2

Complete the square for $x^{2} + 2 x$ .

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Step 1.2.1

Use the form $a x^{2} + b x + c$ , to find the values of $a$ , $b$ , and $c$ .

$a = 1$

$b = 2$

$c = 0$

Step 1.2.2

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e$

Step 1.2.3

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

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Step 1.2.3.1

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{2}{2 \cdot 1}$

Step 1.2.3.2

Cancel the common factor of $2$ .

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Step 1.2.3.2.1

Cancel the common factor.

$d = \frac{2}{2 \cdot 1}$

Step 1.2.3.2.2

Rewrite the expression.

$d = 1$

Step 1.2.4

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

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Step 1.2.4.1

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = 0 - \frac{2^{2}}{4 \cdot 1}$

Step 1.2.4.2

Simplify the right side.

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Step 1.2.4.2.1

Simplify each term.

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Step 1.2.4.2.1.1

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

$e = 0 - \frac{4}{4 \cdot 1}$

Step 1.2.4.2.1.2

Multiply $4$ by $1$ .

$e = 0 - \frac{4}{4}$

Step 1.2.4.2.1.3

Cancel the common factor of $4$ .

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Step 1.2.4.2.1.3.1

Cancel the common factor.

$e = 0 - \frac{4}{4}$

Step 1.2.4.2.1.3.2

Rewrite the expression.

$e = 0 - 1 \cdot 1$

Step 1.2.4.2.1.4

Multiply $- 1$ by $1$ .

$e = 0 - 1$

Step 1.2.4.2.2

Subtract $1$ from $0$ .

$e = - 1$

Step 1.2.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form ${(x + 1)}^{2} - 1$ .

${(x + 1)}^{2} - 1$

Step 1.3

Substitute ${(x + 1)}^{2} - 1$ for $x^{2} + 2 x$ in the equation $x^{2} + 9 y^{2} + 2 x + 18 y = 26$ .

${(x + 1)}^{2} - 1 + 9 y^{2} + 18 y = 26$

Step 1.4

Move $- 1$ to the right side of the equation by adding $1$ to both sides.

${(x + 1)}^{2} + 9 y^{2} + 18 y = 26 + 1$

Step 1.5

Complete the square for $9 y^{2} + 18 y$ .

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Step 1.5.1

Use the form $a x^{2} + b x + c$ , to find the values of $a$ , $b$ , and $c$ .

$a = 9$

$b = 18$

$c = 0$

Step 1.5.2

Consider the vertex form of a parabola.

$a {(x + d)}^{2} + e$

Step 1.5.3

Find the value of $d$ using the formula $d = \frac{b}{2 a}$ .

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Step 1.5.3.1

Substitute the values of $a$ and $b$ into the formula $d = \frac{b}{2 a}$ .

$d = \frac{18}{2 \cdot 9}$

Step 1.5.3.2

Simplify the right side.

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Step 1.5.3.2.1

Cancel the common factor of $18$ and $2$ .

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Step 1.5.3.2.1.1

Factor $2$ out of $18$ .

$d = \frac{2 \cdot 9}{2 \cdot 9}$

Step 1.5.3.2.1.2

Cancel the common factors.

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Step 1.5.3.2.1.2.1

Factor $2$ out of $2 \cdot 9$ .

$d = \frac{2 \cdot 9}{2 (9)}$

Step 1.5.3.2.1.2.2

Cancel the common factor.

$d = \frac{2 \cdot 9}{2 \cdot 9}$

Step 1.5.3.2.1.2.3

Rewrite the expression.

$d = \frac{9}{9}$

Step 1.5.3.2.2

Cancel the common factor of $9$ .

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Step 1.5.3.2.2.1

Cancel the common factor.

$d = \frac{9}{9}$

Step 1.5.3.2.2.2

Rewrite the expression.

$d = 1$

Step 1.5.4

Find the value of $e$ using the formula $e = c - \frac{b^{2}}{4 a}$ .

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Step 1.5.4.1

Substitute the values of $c$ , $b$ and $a$ into the formula $e = c - \frac{b^{2}}{4 a}$ .

$e = 0 - \frac{18^{2}}{4 \cdot 9}$

Step 1.5.4.2

Simplify the right side.

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Step 1.5.4.2.1

Simplify each term.

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Step 1.5.4.2.1.1

Raise $18$ to the power of $2$ .

$e = 0 - \frac{324}{4 \cdot 9}$

Step 1.5.4.2.1.2

Multiply $4$ by $9$ .

$e = 0 - \frac{324}{36}$

Step 1.5.4.2.1.3

Divide $324$ by $36$ .

$e = 0 - 1 \cdot 9$

Step 1.5.4.2.1.4

Multiply $- 1$ by $9$ .

$e = 0 - 9$

Step 1.5.4.2.2

Subtract $9$ from $0$ .

$e = - 9$

Step 1.5.5

Substitute the values of $a$ , $d$ , and $e$ into the vertex form $9 {(y + 1)}^{2} - 9$ .

$9 {(y + 1)}^{2} - 9$

Step 1.6

Substitute $9 {(y + 1)}^{2} - 9$ for $9 y^{2} + 18 y$ in the equation $x^{2} + 9 y^{2} + 2 x + 18 y = 26$ .

${(x + 1)}^{2} + 9 {(y + 1)}^{2} - 9 = 26 + 1$

Step 1.7

Move $- 9$ to the right side of the equation by adding $9$ to both sides.

${(x + 1)}^{2} + 9 {(y + 1)}^{2} = 26 + 1 + 9$

Step 1.8

Simplify $26 + 1 + 9$ .

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Step 1.8.1

Add $26$ and $1$ .

${(x + 1)}^{2} + 9 {(y + 1)}^{2} = 27 + 9$

Step 1.8.2

Add $27$ and $9$ .

${(x + 1)}^{2} + 9 {(y + 1)}^{2} = 36$

Step 1.9

Divide each term by $36$ to make the right side equal to one.

$\frac{{(x + 1)}^{2}}{36} + \frac{9 {(y + 1)}^{2}}{36} = \frac{36}{36}$

Step 1.10

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 $1$ .

$\frac{{(x + 1)}^{2}}{36} + \frac{{(y + 1)}^{2}}{4} = 1$

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.

$a = 6$

$b = 2$

$k = - 1$

$h = - 1$

Step 4

The center of an ellipse follows the form of $(h, k)$ . Substitute in the values of $h$ and $k$ .

$(- 1, - 1)$

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.

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

Step 5.3

Simplify.

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Step 5.3.1

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

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

Step 5.3.2

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

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

Step 5.3.3

Multiply $- 1$ by $4$ .

$\sqrt{36 - 4}$

Step 5.3.4

Subtract $4$ from $36$ .

$\sqrt{32}$

Step 5.3.5

Rewrite $32$ as $4^{2} \cdot 2$ .

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Step 5.3.5.1

Factor $16$ out of $32$ .

$\sqrt{16 (2)}$

Step 5.3.5.2

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

$\sqrt{4^{2} \cdot 2}$

Step 5.3.6

Pull terms out from under the radical.

$4 \sqrt{2}$

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.

$(- 1 + 6, - 1)$

Step 6.3

Simplify.

$(5, - 1)$

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.

$(- 1 - (6), - 1)$

Step 6.6

Simplify.

$(- 7, - 1)$

Step 6.7

Ellipses have two vertices.

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

${Vertex}_{2}$ : $(- 7, - 1)$

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

${Vertex}_{2}$ : $(- 7, - 1)$

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.

$(- 1 + 4 \sqrt{2}, - 1)$

Step 7.3

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

$(h - c, k)$

Step 7.4

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

$(- 1 - (4 \sqrt{2}), - 1)$

Step 7.5

Simplify.

$(- 1 - 4 \sqrt{2}, - 1)$

Step 7.6

Ellipses have two foci.

${Focus}_{1}$ : $(- 1 + 4 \sqrt{2}, - 1)$

${Focus}_{2}$ : $(- 1 - 4 \sqrt{2}, - 1)$

${Focus}_{1}$ : $(- 1 + 4 \sqrt{2}, - 1)$

${Focus}_{2}$ : $(- 1 - 4 \sqrt{2}, - 1)$

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.

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

Step 8.3

Simplify.

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Step 8.3.1

Simplify the numerator.

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Step 8.3.1.1

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

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

Step 8.3.1.2

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

$\frac{\sqrt{36 - 1 \cdot 4}}{6}$

Step 8.3.1.3

Multiply $- 1$ by $4$ .

$\frac{\sqrt{36 - 4}}{6}$

Step 8.3.1.4

Subtract $4$ from $36$ .

$\frac{\sqrt{32}}{6}$

Step 8.3.1.5

Rewrite $32$ as $4^{2} \cdot 2$ .

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Step 8.3.1.5.1

Factor $16$ out of $32$ .

$\frac{\sqrt{16 (2)}}{6}$

Step 8.3.1.5.2

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

$\frac{\sqrt{4^{2} \cdot 2}}{6}$

Step 8.3.1.6

Pull terms out from under the radical.

$\frac{4 \sqrt{2}}{6}$

Step 8.3.2

Cancel the common factor of $4$ and $6$ .

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Step 8.3.2.1

Factor $2$ out of $4 \sqrt{2}$ .

$\frac{2 (2 \sqrt{2})}{6}$

Step 8.3.2.2

Cancel the common factors.

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Step 8.3.2.2.1

Factor $2$ out of $6$ .

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

Step 8.3.2.2.2

Cancel the common factor.

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

Step 8.3.2.2.3

Rewrite the expression.

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

Step 9

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

Center: $(- 1, - 1)$

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

${Vertex}_{2}$ : $(- 7, - 1)$

${Focus}_{1}$ : $(- 1 + 4 \sqrt{2}, - 1)$

${Focus}_{2}$ : $(- 1 - 4 \sqrt{2}, - 1)$

Eccentricity: $\frac{2 \sqrt{2}}{3}$

Step 10